Further Algebra

Further Algebra

Further algebra builds on the polynomial and algebraic techniques from A Level mathematics, extending to partial fractions with irreducible quadratics, the relationships between roots and coefficients of polynomial equations, and systematic summation of series using the method of differences.

Board Coverage

Board	Paper	Notes
AQA	Paper 1	Further partial fractions; roots and coefficients
Edexcel	FP1/FP2	Summation of series; roots of polynomials
OCR (A)	Paper 1	All topics; summation of series emphasised
CIE	P1/P3	Summation of series required; partial fractions in depth

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1. Polynomial Division and the Remainder Theorem

1.1 Polynomial long division

To divide $P(x)$ by $(ax + b)$, perform polynomial long division (or synthetic division) to obtain:

$P(x) = (ax + b)Q(x) + R$

where $Q(x)$ is the quotient and $R$ is a constant remainder.

Proof of the remainder theorem

Proof

Let $P(x)$ be divided by $(x - c)$:

$P(x) = (x - c)Q(x) + R$

for some polynomial $Q(x)$ and constant $R$. Setting $x = c$:

$P(c) = (c - c)Q(c) + R = 0 + R = R$

$\boxed{P(c) = R}$

$\square$

1.2 The factor theorem

Definition. If $P(c) = 0$, then $(x - c)$ is a factor of $P(x)$. This is the factor theorem.

This follows directly from the remainder theorem: if the remainder is zero, the divisor is a factor.

1.3 Finding unknown coefficients

When a polynomial has unknown coefficients, use the factor theorem by substituting known roots, or use the remainder theorem by evaluating at specified points.

Worked Example: Finding unknown coefficients

The polynomial $P(x) = x^3 + ax^2 + bx - 6$ is divisible by $(x - 1)$ and leaves remainder $-24$ when divided by $(x + 3)$. Find $a$ and $b$.

Since $(x - 1)$ is a factor: $P(1) = 1 + a + b - 6 = 0 \implies a + b = 5$ ... (i)

Remainder when divided by $(x + 3)$: $P(-3) = -27 + 9a - 3b - 6 = -24$

$9a - 3b = 9 \implies 3a - b = 3$ ... (ii)

Adding (i) and (ii): $4a = 8 \implies a = 2$. Then $b = 3$.

$P(x) = x^3 + 2x^2 + 3x - 6$.

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2. Partial Fractions with Irreducible Quadratics

In A Level, partial fractions involved only linear factors. In further mathematics, denominators may contain irreducible quadratic factors, requiring a different decomposition.

Definition. A quadratic $x^2 + cx + d$ is irreducible if it has no real roots, i.e. $\Delta = c^2 - 4d < 0$.

2.1 Type 1: Linear times irreducible quadratic

$\boxed{\frac{px + q}{(ax + b)(x^2 + cx + d)} = \frac{A}{ax + b} + \frac{Bx + C}{x^2 + cx + d}}$

The numerator of the irreducible quadratic factor is always linear ($Bx + C$), not just a constant.

Worked Example: Type 1 partial fractions

Express $\dfrac{3x + 5}{(x + 1)(x^2 + 1)}$ in partial fractions.

$\frac{3x + 5}{(x + 1)(x^2 + 1)} = \frac{A}{x + 1} + \frac{Bx + C}{x^2 + 1}$

$3x + 5 = A(x^2 + 1) + (Bx + C)(x + 1)$

Setting $x = -1$: $3(-1) + 5 = A(2) \implies A = 1$.

Setting $x = 0$: $5 = A + C \implies C = 4$.

Setting $x = 1$: $8 = 2A + (B + C)(2) = 2 + 2(B + 4) \implies 2B + 10 = 6 \implies B = -2$.

$\frac{3x + 5}{(x + 1)(x^2 + 1)} = \frac{1}{x + 1} + \frac{-2x + 4}{x^2 + 1}$

2.2 Type 2: Repeated irreducible quadratic

$\boxed{\frac{px^2 + qx + r}{(x^2 + a)^2} = \frac{Ax + B}{x^2 + a} + \frac{Cx + D}{(x^2 + a)^2}}$

When the irreducible quadratic is repeated, the numerators follow the same pattern as repeated linear factors.

2.3 Type 3: Distinct irreducible quadratics

$\boxed{\frac{px^2 + qx + r}{(x^2 + cx + d)(x^2 + ex + f)} = \frac{Ax + B}{x^2 + cx + d} + \frac{Cx + D}{x^2 + ex + f}}$

Each distinct irreducible quadratic factor contributes a linear numerator.

Worked Example: Type 2 partial fractions

Express $\dfrac{x^2 + 1}{(x^2 + 4)^2}$ in partial fractions.

$\frac{x^2 + 1}{(x^2 + 4)^2} = \frac{Ax + B}{x^2 + 4} + \frac{Cx + D}{(x^2 + 4)^2}$

$x^2 + 1 = (Ax + B)(x^2 + 4) + Cx + D = Ax^3 + Bx^2 + 4Ax + 4B + Cx + D$

Comparing coefficients:

• $x^3$: $A = 0$
• $x^2$: $B = 1$
• $x^1$: $4A + C = 0 \implies C = 0$
• $x^0$: $4B + D = 1 \implies 4 + D = 1 \implies D = -3$

$\frac{x^2 + 1}{(x^2 + 4)^2} = \frac{1}{x^2 + 4} - \frac{3}{(x^2 + 4)^2}$

info OCR cover Types 1 and 2. CIE covers Type 1 extensively in P3. :::

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3. Roots of Polynomial Equations

3.1 Cubic equations

If $P(x) = ax^3 + bx^2 + cx + d = a(x - \alpha)(x - \beta)(x - \gamma)$ where $\alpha, \beta, \gamma$ are the roots, then:

$\boxed{\alpha + \beta + \gamma = -\frac{b}{a}}$

$\boxed{\alpha\beta + \alpha\gamma + \beta\gamma = \frac{c}{a}}$

$\boxed{\alpha\beta\gamma = -\frac{d}{a}}$

Proof of the relationship between roots and coefficients for a cubic

Proof

Let $P(x) = ax^3 + bx^2 + cx + d = a(x - \alpha)(x - \beta)(x - \gamma)$.

Expanding the RHS:

$a[(x - \alpha)(x - \beta)(x - \gamma)] = a[x^3 - (\alpha + \beta + \gamma)x^2 + (\alpha\beta + \alpha\gamma + \beta\gamma)x - \alpha\beta\gamma]$

$= ax^3 - a(\alpha + \beta + \gamma)x^2 + a(\alpha\beta + \alpha\gamma + \beta\gamma)x - a\alpha\beta\gamma$

Comparing coefficients with $ax^3 + bx^2 + cx + d$:

• $x^2$: $-a(\alpha + \beta + \gamma) = b \implies \alpha + \beta + \gamma = -\dfrac{b}{a}$
• $x^1$: $a(\alpha\beta + \alpha\gamma + \beta\gamma) = c \implies \alpha\beta + \alpha\gamma + \beta\gamma = \dfrac{c}{a}$
• $x^0$: $-a\alpha\beta\gamma = d \implies \alpha\beta\gamma = -\dfrac{d}{a}$

$\square$

3.2 Quartic equations

For $P(x) = ax^4 + bx^3 + cx^2 + dx + e = a(x - \alpha)(x - \beta)(x - \gamma)(x - \delta)$:

$\boxed{\sum\alpha = \alpha + \beta + \gamma + \delta = -\frac{b}{a}}$

$\boxed{\sum\alpha\beta = \alpha\beta + \alpha\gamma + \alpha\delta + \beta\gamma + \beta\delta + \gamma\delta = \frac{c}{a}}$

$\boxed{\sum\alpha\beta\gamma = -\frac{d}{a}}$

$\boxed{\alpha\beta\gamma\delta = \frac{e}{a}}$

3.3 Symmetric functions of roots

Using the elementary symmetric sums, we can express other symmetric functions:

• $\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \alpha\gamma + \beta\gamma)$
• $\dfrac◆LB◆1◆RB◆◆LB◆\alpha◆RB◆ + \dfrac◆LB◆1◆RB◆◆LB◆\beta◆RB◆ + \dfrac◆LB◆1◆RB◆◆LB◆\gamma◆RB◆ = \dfrac◆LB◆\alpha\beta + \alpha\gamma + \beta\gamma◆RB◆◆LB◆\alpha\beta\gamma◆RB◆$
• $\alpha^2\beta + \alpha^2\gamma + \beta^2\alpha + \beta^2\gamma + \gamma^2\alpha + \gamma^2\beta = (\alpha + \beta + \gamma)(\alpha\beta + \alpha\gamma + \beta\gamma) - 3\alpha\beta\gamma$

Worked Example: Symmetric functions of roots

The equation $2x^3 - 3x^2 - 4x + 5 = 0$ has roots $\alpha, \beta, \gamma$. Find the value of $\alpha^2 + \beta^2 + \gamma^2$.

From the relationships: $\alpha + \beta + \gamma = \dfrac{3}{2}$ and $\alpha\beta + \alpha\gamma + \beta\gamma = \dfrac{-4}{2} = -2$.

$\alpha^2 + \beta^2 + \gamma^2 = \left(\frac{3}{2}\right)^2 - 2(-2) = \frac{9}{4} + 4 = \frac{25}{4}$

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4. Summation of Series

4.1 Standard results

The following summation formulae are essential:

$\boxed{\sum_{r=1}^{n} r = \frac{n(n+1)}{2}}$

$\boxed{\sum_{r=1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6}}$

$\boxed{\sum_{r=1}^{n} r^3 = \left[\frac{n(n+1)}{2}\right]^2}$

4.2 The method of differences

To find $\displaystyle\sum_{r=1}^{n} f(r)$ where $f(r)$ can be written as $g(r) - g(r+1)$:

$\sum_{r=1}^{n} f(r) = \sum_{r=1}^{n} [g(r) - g(r+1)] = g(1) - g(n+1)$

This is a telescoping sum — all intermediate terms cancel.

Proof of the sum of squares formula by the method of differences

Proof

Note that $r^3 - (r-1)^3 = 3r^2 - 3r + 1$, so $r^2 = \dfrac{r^3 - (r-1)^3 + 3r - 1}{3}$.

Summing from $r = 1$ to $n$:

$\sum_{r=1}^{n} r^2 = \frac{1}{3}\sum_{r=1}^{n}[r^3 - (r-1)^3] + \sum_{r=1}^{n} r - \frac{n}{3}$

The first sum telescopes: $\sum_{r=1}^{n}[r^3 - (r-1)^3] = n^3 - 0 = n^3$.

$\sum_{r=1}^{n} r^2 = \frac{n^3}{3} + \frac{n(n+1)}{2} - \frac{n}{3} = \frac{2n^3 + 3n^2 + 3n + 2n^2 + 2n - 2n}{6} \cdot \frac{1}{1}$

More carefully:

$\sum_{r=1}^{n} r^2 = \frac{n^3}{3} + \frac{n(n+1)}{2} - \frac{n}{3} = \frac{2n^3 + 3n(n+1) - 2n}{6} = \frac{2n^3 + 3n^2 + 3n - 2n}{6}$

$= \frac{2n^3 + 3n^2 + n}{6} = \frac{n(2n^2 + 3n + 1)}{6} = \frac{n(n+1)(2n+1)}{6} \quad \square$

4.3 Further standard results

$\boxed{\sum_{r=1}^{n} r(r+1) = \frac{n(n+1)(n+2)}{3}}$

$\boxed{\sum_{r=1}^{n} r(r+1)(r+2) = \frac{n(n+1)(n+2)(n+3)}{4}}$

In general, $\displaystyle\sum_{r=1}^{n} \binom{r+k}{k+1} = \binom{n+k+1}{k+2}$.

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5. Further Summation Techniques

5.1 Using partial fractions for summation

When the general term of a series can be decomposed into partial fractions, the method of differences often applies.

Worked Example: Sum using method of differences

Find $\displaystyle\sum_{r=1}^{n} \frac{1}{r(r+1)}$.

Using partial fractions: $\dfrac{1}{r(r+1)} = \dfrac{1}{r} - \dfrac{1}{r+1}$.

$\sum_{r=1}^{n}\left(\frac{1}{r} - \frac{1}{r+1}\right) = \left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \cdots + \left(\frac{1}{n} - \frac{1}{n+1}\right)$

$= 1 - \frac{1}{n+1} = \frac{n}{n+1}$

Worked Example: Sum with quadratic denominator

Find $\displaystyle\sum_{r=1}^{n} \frac{1}{r(r+2)}$.

Partial fractions: $\dfrac{1}{r(r+2)} = \dfrac{1}{2}\!\left(\dfrac{1}{r} - \dfrac{1}{r+2}\right)$.

$\frac{1}{2}\sum_{r=1}^{n}\left(\frac{1}{r} - \frac{1}{r+2}\right) = \frac{1}{2}\left[\left(1 - \frac{1}{3}\right) + \left(\frac{1}{2} - \frac{1}{4}\right) + \left(\frac{1}{3} - \frac{1}{5}\right) + \cdots + \left(\frac{1}{n} - \frac{1}{n+2}\right)\right]$

Terms cancel in pairs. The surviving terms are $1 + \dfrac{1}{2} - \dfrac{1}{n+1} - \dfrac{1}{n+2}$.

$= \frac{1}{2}\left(\frac{3}{2} - \frac{1}{n+1} - \frac{1}{n+2}\right) = \frac{3}{4} - \frac{2n+3}{2(n+1)(n+2)}$

5.2 Summation of $r \cdot a_r$

To find $\displaystyle\sum_{r=1}^{n} r \cdot a_r$ where $a_r = f(r) - f(r-1)$:

$\sum_{r=1}^{n} r \cdot a_r = \sum_{r=1}^{n} r[f(r) - f(r-1)] = nf(n) - \sum_{r=0}^{n-1} f(r)$

This is known as the summation by parts technique.

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6. Proofs of Standard Summation Formulae

Proof of $\sum r = \frac{n(n+1)}{2}$ (by induction)

Proof

Base case ($n = 1$): $\displaystyle\sum_{r=1}^{1} r = 1 = \frac◆LB◆1 \times 2◆RB◆◆LB◆2◆RB◆$. ✓

Inductive step. Assume $\displaystyle\sum_{r=1}^{k} r = \frac{k(k+1)}{2}$. Then:

$\sum_{r=1}^{k+1} r = \frac{k(k+1)}{2} + (k+1) = \frac{k(k+1) + 2(k+1)}{2} = \frac{(k+1)(k+2)}{2}$

✓ $\square$

Proof of $\sum r^3 = \left[\frac{n(n+1)}{2}\right]^2$ (by induction)

Proof

Base case ($n = 1$): $1^3 = 1 = \left[\dfrac◆LB◆1 \times 2◆RB◆◆LB◆2◆RB◆\right]^2 = 1$. ✓

Inductive step. Assume $\displaystyle\sum_{r=1}^{k} r^3 = \left[\frac{k(k+1)}{2}\right]^2$. Then:

$\sum_{r=1}^{k+1} r^3 = \left[\frac{k(k+1)}{2}\right]^2 + (k+1)^3 = \frac{k^2(k+1)^2}{4} + \frac{4(k+1)^3}{4}$

$= \frac{(k+1)^2[k^2 + 4(k+1)]}{4} = \frac{(k+1)^2(k+2)^2}{4} = \left[\frac{(k+1)(k+2)}{2}\right]^2$

✓ $\square$

info summation formulae. Edexcel FP2 requires summation of series including method of

differences. AQA covers summation in the context of mathematical induction. :::

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7. Summary of Key Results

$\boxed{P(c) = R \quad \mathrm{(Remainder Theorem)}}$

$\boxed{\frac{px + q}{(ax + b)(x^2 + cx + d)} = \frac{A}{ax + b} + \frac{Bx + C}{x^2 + cx + d}}$

$\boxed{\alpha + \beta + \gamma = -\frac{b}{a}, \quad \alpha\beta + \alpha\gamma + \beta\gamma = \frac{c}{a}, \quad \alpha\beta\gamma = -\frac{d}{a}}$

$\boxed{\sum_{r=1}^{n} r = \frac{n(n+1)}{2}, \quad \sum_{r=1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6}, \quad \sum_{r=1}^{n} r^3 = \left[\frac{n(n+1)}{2}\right]^2}$

$\boxed{\sum_{r=1}^{n} [g(r) - g(r+1)] = g(1) - g(n+1)}$

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Problems

Problem 1. Express $\dfrac{2x^2 + 3x + 4}{(x + 2)(x^2 + 2x + 5)}$ in partial fractions.

Hint

Since $x^2 + 2x + 5 = (x+1)^2 + 4$ has $\Delta = 4 - 20 < 0$, it is irreducible. Use the form $\dfrac{A}{x+2} + \dfrac{Bx + C}{x^2 + 2x + 5}$.

Answer

$\frac{2x^2 + 3x + 4}{(x + 2)(x^2 + 2x + 5)} = \frac{A}{x + 2} + \frac{Bx + C}{x^2 + 2x + 5}$

$2x^2 + 3x + 4 = A(x^2 + 2x + 5) + (Bx + C)(x + 2)$

Setting $x = -2$: $8 - 6 + 4 = A(4 + 1) = 5A \implies A = \dfrac{6}{5}$.

Setting $x = 0$: $4 = 5A + 2C = 6 + 2C \implies C = -1$.

Setting $x = 1$: $2 + 3 + 4 = 5A + (B - 1)(3) = 6 + 3B - 3 \implies 9 = 3 + 3B \implies B = 2$.

$\frac{2x^2 + 3x + 4}{(x + 2)(x^2 + 2x + 5)} = \frac{6/5}{x + 2} + \frac{2x - 1}{x^2 + 2x + 5}$

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Problem 2. The equation $x^3 - 4x^2 + x + 6 = 0$ has roots $\alpha, \beta, \gamma$. Find the value of $\dfrac◆LB◆1◆RB◆◆LB◆\alpha\beta◆RB◆ + \dfrac◆LB◆1◆RB◆◆LB◆\alpha\gamma◆RB◆ + \dfrac◆LB◆1◆RB◆◆LB◆\beta\gamma◆RB◆$.

Hint

$\dfrac◆LB◆1◆RB◆◆LB◆\alpha\beta◆RB◆ + \dfrac◆LB◆1◆RB◆◆LB◆\alpha\gamma◆RB◆ + \dfrac◆LB◆1◆RB◆◆LB◆\beta\gamma◆RB◆ = \dfrac◆LB◆\alpha + \beta + \gamma◆RB◆◆LB◆\alpha\beta\gamma◆RB◆$.

Answer

$\alpha + \beta + \gamma = \dfrac{-(-4)}{1} = 4$ and $\alpha\beta\gamma = \dfrac{-6}{1} = -6$.

$\frac◆LB◆1◆RB◆◆LB◆\alpha\beta◆RB◆ + \frac◆LB◆1◆RB◆◆LB◆\alpha\gamma◆RB◆ + \frac◆LB◆1◆RB◆◆LB◆\beta\gamma◆RB◆ = \frac◆LB◆\alpha + \beta + \gamma◆RB◆◆LB◆\alpha\beta\gamma◆RB◆ = \frac{4}{-6} = -\frac{2}{3}$

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Problem 3. Express $\dfrac{3x + 1}{(x^2 + 1)(x^2 + 4)}$ in partial fractions.

Hint

Both $x^2 + 1$ and $x^2 + 4$ are irreducible. Use the form $\dfrac{Ax + B}{x^2 + 1} + \dfrac{Cx + D}{x^2 + 4}$.

Answer

$\frac{3x + 1}{(x^2 + 1)(x^2 + 4)} = \frac{Ax + B}{x^2 + 1} + \frac{Cx + D}{x^2 + 4}$

$3x + 1 = (Ax + B)(x^2 + 4) + (Cx + D)(x^2 + 1)$

$= (A + C)x^3 + (B + D)x^2 + (4A + C)x + (4B + D)$

Comparing coefficients:

• $x^3$: $A + C = 0$
• $x^2$: $B + D = 0$
• $x^1$: $4A + C = 3$
• $x^0$: $4B + D = 1$

From $A + C = 0$ and $4A + C = 3$: $3A = 3 \implies A = 1, C = -1$.

From $B + D = 0$ and $4B + D = 1$: $3B = 1 \implies B = \dfrac{1}{3}, D = -\dfrac{1}{3}$.

$\frac{3x + 1}{(x^2 + 1)(x^2 + 4)} = \frac{x + 1/3}{x^2 + 1} + \frac{-x - 1/3}{x^2 + 4}$

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Problem 4. Find $\displaystyle\sum_{r=1}^{n} \frac{2}{r(r+1)(r+2)}$.

Hint

Use partial fractions to show that $\dfrac{2}{r(r+1)(r+2)} = \dfrac{1}{r(r+1)} - \dfrac{1}{(r+1)(r+2)}$, then apply the method of differences.

Answer

$\dfrac{2}{r(r+1)(r+2)} = \dfrac{1}{r(r+1)} - \dfrac{1}{(r+1)(r+2)}$.

This telescopes:

$\sum_{r=1}^{n}\left[\frac{1}{r(r+1)} - \frac{1}{(r+1)(r+2)}\right] = \frac◆LB◆1◆RB◆◆LB◆1 \times 2◆RB◆ - \frac{1}{(n+1)(n+2)}$

$= \frac{1}{2} - \frac{1}{(n+1)(n+2)}$

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Problem 5. The equation $3x^3 + px^2 + qx + 12 = 0$ has roots $\alpha, \beta, \gamma$ such that $\alpha + \beta + \gamma = 4$ and $\alpha\beta\gamma = -4$. Find $p$, $q$, and $\alpha\beta + \alpha\gamma + \beta\gamma$.

Hint

Use the relationships between roots and coefficients directly.

Answer

$\alpha + \beta + \gamma = -\dfrac{p}{3} = 4 \implies p = -12$.

$\alpha\beta\gamma = -\dfrac{12}{3} = -4$. This is consistent with the given information. ✓

$\alpha\beta + \alpha\gamma + \beta\gamma = \dfrac{q}{3}$, so $q = 3(\alpha\beta + \alpha\gamma + \beta\gamma)$.

We need additional information. Note that $\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \alpha\gamma + \beta\gamma)$ $= 16 - 2S$ where $S = \alpha\beta + \alpha\gamma + \beta\gamma$.

Without further information about the individual roots, $S$ cannot be determined uniquely. However, we know $p = -12$ and $q$ depends on $S$.

If the question provides that the roots are integers: trying factors of $\dfrac{-4}{3}$, the roots are $1, 1, 2$ (checking: sum = 4 ✓, product = 2 ≠ $-4$ ✗). The roots $-1, 2, 3$ give sum = 4 ✓ and product = $-6$ ✗.

$p = -12$ and $q = 3S$ where $S$ requires more information about the roots.

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Problem 6. Find $\displaystyle\sum_{r=1}^{n} \frac{1}{r(r+3)}$.

Hint

Use partial fractions: $\dfrac{1}{r(r+3)} = \dfrac{1}{3}\!\left(\dfrac{1}{r} - \dfrac{1}{r+3}\right)$. Three terms survive the telescoping.

Answer

$\frac{1}{3}\sum_{r=1}^{n}\left(\frac{1}{r} - \frac{1}{r+3}\right) = \frac{1}{3}\left[\left(1 - \frac{1}{4}\right) + \left(\frac{1}{2} - \frac{1}{5}\right) + \left(\frac{1}{3} - \frac{1}{6}\right) + \cdots + \left(\frac{1}{n} - \frac{1}{n+3}\right)\right]$

The surviving terms are $\dfrac{1}{1} + \dfrac{1}{2} + \dfrac{1}{3} - \dfrac{1}{n+1} - \dfrac{1}{n+2} - \dfrac{1}{n+3}$.

$= \frac{1}{3}\left(\frac{11}{6} - \frac{1}{n+1} - \frac{1}{n+2} - \frac{1}{n+3}\right) = \frac{11}{18} - \frac{1}{3}\!\left(\frac{1}{n+1} + \frac{1}{n+2} + \frac{1}{n+3}\right)$

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Problem 7. The polynomial $P(x) = x^4 + ax^3 + bx^2 + cx + d$ has roots $\alpha, \beta, \gamma, \delta$. Given that $\alpha + \beta = 3$, $\gamma + \delta = -5$, and $\alpha\beta = 2$, find $a$ and $b$.

Hint

Use $\sum\alpha = -\dfrac{a}{1}$ and $\sum\alpha\beta = \dfrac{b}{1}$.

Answer

$\sum\alpha = \alpha + \beta + \gamma + \delta = 3 + (-5) = -2$.

$a = -\sum\alpha = 2$.

$\sum\alpha\beta = \alpha\beta + \alpha\gamma + \alpha\delta + \beta\gamma + \beta\delta + \gamma\delta$.

$= \alpha\beta + (\alpha + \beta)(\gamma + \delta) + \gamma\delta = 2 + (3)(-5) + \gamma\delta = 2 - 15 + \gamma\delta = -13 + \gamma\delta$.

We need $\gamma\delta$. Since we don't have $\gamma\delta$ directly, $b = -13 + \gamma\delta$.

$a = 2$ and $b$ depends on $\gamma\delta$ (which requires further information to determine).

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Problem 8. Prove by induction that $\displaystyle\sum_{r=1}^{n} r(r+1) = \frac{n(n+1)(n+2)}{3}$ for all $n \in \mathbb{Z}^+$.

Hint

Base case: $n = 1$. Inductive step: assume for $n = k$ and add the $(k+1)$-th term.

Answer

Base case ($n = 1$): $1 \times 2 = 2 = \dfrac◆LB◆1 \times 2 \times 3◆RB◆◆LB◆3◆RB◆ = 2$. ✓

Inductive step. Assume $\displaystyle\sum_{r=1}^{k} r(r+1) = \frac{k(k+1)(k+2)}{3}$. Then:

$\sum_{r=1}^{k+1} r(r+1) = \frac{k(k+1)(k+2)}{3} + (k+1)(k+2)$

$= \frac{(k+1)(k+2)[k + 3]}{3} = \frac{(k+1)(k+2)(k+3)}{3}$

✓ $\square$

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Problem 9. Express $\dfrac{x^2 + 3x + 2}{(x^2 + 2x + 3)^2}$ in partial fractions.

Hint

Use the form $\dfrac{Ax + B}{x^2 + 2x + 3} + \dfrac{Cx + D}{(x^2 + 2x + 3)^2}$.

Answer

$x^2 + 3x + 2 = (Ax + B)(x^2 + 2x + 3) + Cx + D$

$= Ax^3 + (2A + B)x^2 + (3A + 2B + C)x + (3B + D)$

Comparing coefficients:

• $x^3$: $A = 0$
• $x^2$: $B = 1$
• $x^1$: $2 + C = 3 \implies C = 1$
• $x^0$: $3 + D = 2 \implies D = -1$

$\frac{x^2 + 3x + 2}{(x^2 + 2x + 3)^2} = \frac{1}{x^2 + 2x + 3} + \frac{x - 1}{(x^2 + 2x + 3)^2}$

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Problem 10. The cubic equation $x^3 + px^2 + qx + r = 0$ has roots $\alpha, \beta, \gamma$ where $\beta = 2\alpha$ and $\gamma = 3\alpha$. Express $p$, $q$, and $r$ in terms of $\alpha$, and hence find the roots when $p = -6$.

Hint

Substitute the root relationships into $\alpha + \beta + \gamma = -p$, $\alpha\beta + \alpha\gamma + \beta\gamma = q$, and $\alpha\beta\gamma = -r$.

Answer

$\alpha + 2\alpha + 3\alpha = 6\alpha = -p$, so $p = -6\alpha$.

$\alpha(2\alpha) + \alpha(3\alpha) + (2\alpha)(3\alpha) = 2\alpha^2 + 3\alpha^2 + 6\alpha^2 = 11\alpha^2 = q$.

$\alpha(2\alpha)(3\alpha) = 6\alpha^3 = -r$, so $r = -6\alpha^3$.

When $p = -6$: $-6\alpha = -6 \implies \alpha = 1$.

Then $q = 11$, $r = -6$, and the roots are $1, 2, 3$.

Verification: $(x-1)(x-2)(x-3) = x^3 - 6x^2 + 11x - 6$. ✓

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8. Advanced Worked Examples

Example 8.1: Using the binomial theorem with negative and fractional indices

Problem. Find the coefficient of $x^4$ in the expansion of $(1 - 2x)^{-1/2}$ up to and including the term in $x^4$.

Solution. Using the general binomial expansion for $|x| < \dfrac{1}{2}$:

$(1+y)^n = 1 + ny + \frac{n(n-1)}{2!}y^2 + \frac{n(n-1)(n-2)}{3!}y^3 + \frac{n(n-1)(n-2)(n-3)}{4!}y^4 + \cdots$

With $n = -\dfrac{1}{2}$ and $y = -2x$:

$(1-2x)^{-1/2} = 1 + \left(-\frac{1}{2}\right)(-2x) + \frac◆LB◆\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)◆RB◆◆LB◆2◆RB◆(-2x)^2 + \cdots$

$= 1 + x + \frac{3}{8}(4x^2) + \cdots = 1 + x + \frac{3}{2}x^2 + \cdots$

The $x^4$ coefficient: $\dfrac◆LB◆\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)\left(-\frac{7}{2}\right)◆RB◆◆LB◆24◆RB◆(16) = \dfrac{105}{16} \cdot \dfrac{16}{24} = \dfrac{105}{24} = \boxed{\dfrac{35}{8}}$.

Example 8.2: Roots of a cubic with a substitution

Problem. The cubic equation $x^3 - 3x^2 + 4 = 0$ has roots $\alpha, \beta, \gamma$. Find the value of $\alpha^2 + \beta^2 + \gamma^2$.

Solution. By Vieta's formulae: $\alpha + \beta + \gamma = 3$ and $\alpha\beta + \beta\gamma + \gamma\alpha = 0$.

$\alpha^2 + \beta^2 + \gamma^2 = (\alpha+\beta+\gamma)^2 - 2(\alpha\beta+\beta\gamma+\gamma\alpha) = 9 - 0 = \boxed{9}$

Example 8.3: Telescoping series via partial fractions

Problem. Find $\displaystyle\sum_{r=1}^{n} \frac{1}{r(r+1)}$ and deduce $\displaystyle\sum_{r=1}^{\infty} \frac{1}{r(r+1)}$.

Solution. $\dfrac{1}{r(r+1)} = \dfrac{1}{r} - \dfrac{1}{r+1}$.

$\sum_{r=1}^{n} \frac{1}{r(r+1)} = \left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \cdots + \left(\frac{1}{n} - \frac{1}{n+1}\right) = 1 - \frac{1}{n+1} = \frac{n}{n+1}$

As $n \to \infty$: $\displaystyle\sum_{r=1}^{\infty} \frac{1}{r(r+1)} = \boxed{1}$.

Example 8.4: Proof by induction on a binomial coefficient identity

Problem. Prove by induction that $\displaystyle\sum_{r=1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6}$.

Solution. Base case ($n=1$): LHS $= 1$, RHS $= \dfrac◆LB◆1 \cdot 2 \cdot 3◆RB◆◆LB◆6◆RB◆ = 1$. ✓

Inductive hypothesis: Assume $\displaystyle\sum_{r=1}^{k} r^2 = \frac{k(k+1)(2k+1)}{6}$.

Inductive step: $\displaystyle\sum_{r=1}^{k+1} r^2 = \frac{k(k+1)(2k+1)}{6} + (k+1)^2$

$= \frac{k(k+1)(2k+1) + 6(k+1)^2}{6} = \frac{(k+1)[k(2k+1) + 6(k+1)]}{6}$

$= \frac{(k+1)(2k^2 + k + 6k + 6)}{6} = \frac{(k+1)(2k^2 + 7k + 6)}{6} = \frac{(k+1)(k+2)(2k+3)}{6}$

This is the required form with $n = k+1$. $\blacksquare$

Example 8.5: Method of differences with rational expressions

Problem. Find $\displaystyle\sum_{r=1}^{n} \frac{1}{(2r-1)(2r+1)}$.

Solution. $\dfrac{1}{(2r-1)(2r+1)} = \dfrac{1}{2}\!\left(\dfrac{1}{2r-1} - \dfrac{1}{2r+1}\right)$.

$\sum_{r=1}^{n} \frac{1}{(2r-1)(2r+1)} = \frac{1}{2}\left[\left(\frac{1}{1} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{5}\right) + \cdots + \left(\frac{1}{2n-1} - \frac{1}{2n+1}\right)\right]$

$= \frac{1}{2}\left(1 - \frac{1}{2n+1}\right) = \boxed{\frac{n}{2n+1}}$

Example 8.6: Manipulating series with a given recurrence

Problem. Given $u_1 = 1$ and $u_{n+1} = \dfrac{u_n}{u_n + 1}$, find $\displaystyle\sum_{r=1}^{n} u_r$.

Solution. Write $u_r$ in closed form. From the recurrence: $\dfrac{1}{u_{n+1}} = \dfrac{u_n + 1}{u_n} = 1 + \dfrac{1}{u_n}$.

Let $v_n = \dfrac{1}{u_n}$. Then $v_{n+1} = 1 + v_n$, so $v_n = v_1 + (n-1)$.

Since $v_1 = \dfrac{1}{u_1} = 1$: $v_n = n$, so $u_n = \dfrac{1}{n}$.

$\sum_{r=1}^{n} u_r = \sum_{r=1}^{n} \frac{1}{r} = H_n$

The $n$-th harmonic number. No simpler closed form exists.

Example 8.7: Simultaneous equations via matrices

Problem. Solve the system $x + 2y - z = 3$, $2x - y + z = 4$, $3x + y + 2z = 7$.

Solution. In matrix form $\mathbf{M}\mathbf{x} = \mathbf{b}$:

$\mathbf{M} = \begin{pmatrix}1&2&-1\\2&-1&1\\3&1&2\end{pmatrix}$

$\det(\mathbf{M}) = 1(2-1) - 2(4-3) + (-1)(2+3) = 1 - 2 - 5 = -6 \neq 0$, so the system has a unique solution.

Using Cramer's rule: $x = \dfrac◆LB◆\det\begin{pmatrix}3&2&-1\\4&-1&1\\7&1&2\end{pmatrix}◆RB◆◆LB◆-6◆RB◆ = \dfrac{-6+16-11}{-6} = \dfrac{-1}{-6} = \dfrac{1}{6}$.

Similarly: $y = \dfrac{5}{3}$ and $z = \dfrac{1}{6}$.

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9. Common Pitfalls

Pitfall	Correct Approach
Forgetting the condition $\|x\| < 1$ for binomial expansions	Always state the convergence condition explicitly
Confusing $\displaystyle\sum_{r=1}^{n} r^3$ with $\left(\displaystyle\sum_{r=1}^{n} r\right)^3$	$\sum r^3 = \dfrac{n^2(n+1)^2}{4}$; they happen to be equal but the reasoning is different
Splitting partial fractions incorrectly for method of differences	Always check by recombining: $\dfrac{A}{r} + \dfrac{B}{r+1} = \dfrac{A(r+1) + Br}{r(r+1)}$
Assuming Vieta's formulae give $\alpha\beta\gamma = -d/a$ without checking the sign	For $ax^3+bx^2+cx+d=0$: $\alpha+\beta+\gamma=-b/a$, $\alpha\beta+\beta\gamma+\gamma\alpha=c/a$, $\alpha\beta\gamma=-d/a$
Skipping the base case in induction proofs	The base case is essential — without it the induction chain is unanchored

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10. Additional Exam-Style Questions

Question 8

The binomial expansion of $(1 + ax)^{-2}$, in ascending powers of $x$ up to and including the term in $x^3$, is $1 - 4x + 12x^2 + bx^3$. Find the values of $a$ and $b$.

Solution

$(1+ax)^{-2} = 1 + (-2)(ax) + \dfrac{(-2)(-3)}{2}(ax)^2 + \dfrac{(-2)(-3)(-4)}{6}(ax)^3 + \cdots$

$= 1 - 2ax + 3a^2x^2 - 4a^3x^3 + \cdots$

Comparing: $-2a = -4 \implies a = 2$. Then $b = -4(8) = -32$.

$\boxed{a = 2, \; b = -32}$

Question 9

Prove by induction that $7^n - 1$ is divisible by $6$ for all positive integers $n$.

Solution

Base case ($n=1$): $7^1 - 1 = 6$, divisible by 6. ✓

Inductive hypothesis: $7^k - 1 = 6m$ for some integer $m$.

Inductive step: $7^{k+1} - 1 = 7 \cdot 7^k - 1 = 7(6m + 1) - 1 = 42m + 6 = 6(7m + 1)$.

This is divisible by 6. $\blacksquare$

Question 10

Find $\displaystyle\sum_{r=1}^{n} \frac{2}{r(r+2)}$.

Solution

$\dfrac{2}{r(r+2)} = \dfrac{1}{r} - \dfrac{1}{r+2}$.

$\sum_{r=1}^{n} \frac{2}{r(r+2)} = \left(1 - \frac{1}{3}\right) + \left(\frac{1}{2} - \frac{1}{4}\right) + \left(\frac{1}{3} - \frac{1}{5}\right) + \cdots + \left(\frac{1}{n} - \frac{1}{n+2}\right)$

Terms $\dfrac{1}{3}$ to $\dfrac{1}{n}$ cancel, leaving:

$= 1 + \frac{1}{2} - \frac{1}{n+1} - \frac{1}{n+2} = \frac{3}{2} - \frac{2n+3}{(n+1)(n+2)} = \frac{3(n+1)(n+2) - 2(2n+3)}{2(n+1)(n+2)}$

$= \frac{3n^2 + 9n + 6 - 4n - 6}{2(n+1)(n+2)} = \boxed{\frac{3n^2 + 5n}{2(n+1)(n+2)}}$

Question 11

The equation $x^3 + px^2 + qx + r = 0$ has roots $\alpha, 2\alpha, 3\alpha$. Find $p:q:r$.

Solution

$\alpha + 2\alpha + 3\alpha = -p \implies p = -6\alpha$.

$\alpha(2\alpha) + 2\alpha(3\alpha) + 3\alpha(\alpha) = q \implies 2\alpha^2 + 6\alpha^2 + 3\alpha^2 = 11\alpha^2 = q$.

$\alpha(2\alpha)(3\alpha) = -r \implies 6\alpha^3 = -r$.

$p:q:r = -6\alpha : 11\alpha^2 : -6\alpha^3 = -6 : 11\alpha : -6\alpha^2$.

For specific values, if $\alpha = 1$: $p:q:r = -6:11:-6$, giving $(x-1)(x-2)(x-3) = x^3 - 6x^2 + 11x - 6$.

Question 12

Use the Maclaurin expansion of $(1+x)^{1/2}$ to find $\sqrt{1.02}$ correct to 6 decimal places.

Solution

$(1+x)^{1/2} = 1 + \dfrac{1}{2}x - \dfrac{1}{8}x^2 + \dfrac{1}{16}x^3 - \dfrac{5}{128}x^4 + \cdots$

With $x = 0.02$:

$\sqrt{1.02} = 1 + 0.01 - \dfrac{0.0004}{8} + \dfrac{0.000008}{16} - \dfrac{5(0.02)^4}{128} + \cdots$

$= 1 + 0.01 - 0.00005 + 0.0000005 - 0.000000005 + \cdots = 1.009950495...$

$\boxed{\sqrt{1.02} \approx 1.009950}$

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11. Connections to Other Topics

11.1 Further algebra and complex numbers

The roots of unity and De Moivre's theorem connect algebra to complex numbers. See Complex Numbers.

11.2 Algebra and matrices

Vieta's formulae are closely related to the characteristic equation of a matrix: the sum of eigenvalues equals the trace, and the product equals the determinant. See Matrices.

11.3 Series and calculus

The binomial expansion and Maclaurin series are both infinite series representations of functions, used extensively in integration and differentiation. See Maclaurin and Taylor Series.

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12. Key Formulae Summary

Topic	Key Formula
General binomial	$(1+x)^n = \displaystyle\sum_{k=0}^{\infty} \binom{n}{k}x^k$ for $\|x\| < 1$
Method of differences	Decompose $\dfrac{P(r)}{Q(r)}$ into partial fractions that telescope
Induction	Base case $\to$ assume $P(k)$ $\to$ prove $P(k+1)$
Vieta's (cubic)	$\alpha+\beta+\gamma=-b/a$, $\alpha\beta+\beta\gamma+\gamma\alpha=c/a$, $\alpha\beta\gamma=-d/a$
Sum of squares	$\displaystyle\sum_{r=1}^{n} r^2 = \dfrac{n(n+1)(2n+1)}{6}$
Sum of cubes	$\displaystyle\sum_{r=1}^{n} r^3 = \dfrac{n^2(n+1)^2}{4}$
Harmonic sum	$H_n = \displaystyle\sum_{r=1}^{n} \dfrac{1}{r} \approx \ln n + \gamma$

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13. Further Exam-Style Questions

Question 13

Prove by induction that $3^n > n^3$ for all integers $n \geq 4$.

Solution

Base case ($n=4$): $3^4 = 81 > 64 = 4^3$. ✓

Inductive hypothesis: $3^k > k^3$ for $k \geq 4$.

Inductive step: $3^{k+1} = 3 \cdot 3^k > 3k^3$.

We need $3k^3 > (k+1)^3 = k^3 + 3k^2 + 3k + 1$.

$2k^3 - 3k^2 - 3k - 1 > 0$ for $k \geq 4$.

At $k=4$: $128-48-12-1 = 67 > 0$. ✓

For $k > 4$: $2k^3$ grows faster than $3k^2 + 3k + 1$, so the inequality holds. $\blacksquare$

Question 14

Find the coefficient of $x^3$ in the expansion of $\dfrac{1}{(1-2x)(1+x)}$.

Solution

Partial fractions: $\dfrac{1}{(1-2x)(1+x)} = \dfrac{A}{1-2x} + \dfrac{B}{1+x}$.

$1 = A(1+x) + B(1-2x)$. $x=-1$: $1 = 3A \implies A = 1/3$. $x=1/2$: $1 = \dfrac{3B}{2} \implies B = 2/3$.

$\dfrac{1}{3}\sum (2x)^n + \dfrac{2}{3}\sum (-x)^n$.

$x^3$ coefficient: $\dfrac{1}{3} \cdot 8 + \dfrac{2}{3}(-1) = \dfrac{8-2}{3} = \boxed{2}$.

Question 15

The roots of $x^3 + px + q = 0$ are $\alpha, \beta, \gamma$. Express $\alpha^3 + \beta^3 + \gamma^3$ in terms of $p$ and $q$.

Solution

Since $\alpha$ is a root: $\alpha^3 = -p\alpha - q$. Similarly for $\beta, \gamma$.

$\alpha^3 + \beta^3 + \gamma^3 = -p(\alpha+\beta+\gamma) - 3q$.

For $x^3 + px + q = 0$ (no $x^2$ term): $\alpha+\beta+\gamma = 0$.

$\alpha^3 + \beta^3 + \gamma^3 = \boxed{-3q}$.

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14. Advanced Topics

14.1 The general binomial theorem for any index

For $|x| < 1$ and any real $n$:

$(1+x)^n = \sum_{k=0}^{\infty} \binom{n}{k}x^k = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \cdots$

When $n$ is a positive integer, this terminates at $k = n$. Otherwise, it is an infinite series.

14.2 Summation by parts (discrete integration by parts)

Analogous to integration by parts:

$\sum_{r=a}^{b} u_r \Delta v_r = [u_r v_r]_a^{b+1} - \sum_{r=a}^{b} (\Delta u_r) v_{r+1}$

where $\Delta f(r) = f(r+1) - f(r)$ is the forward difference operator.

14.3 Generating functions

A generating function for a sequence $\{a_n\}$ is $G(x) = \displaystyle\sum_{n=0}^{\infty} a_n x^n$.

Examples:

• $1, 1, 1, \ldots$: $G(x) = \dfrac{1}{1-x}$
• $1, 2, 3, \ldots$: $G(x) = \dfrac{1}{(1-x)^2}$
• $1, 3, 6, 10, \ldots$ (triangular): $G(x) = \dfrac{1}{(1-x)^3}$

14.4 Relationship between binomial coefficients and Pascal's triangle

$\binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}$ (Pascal's identity).

This is the basis of Pascal's triangle and is proved combinatorially: choosing $k$ objects from $n$ either includes or excludes a specific object.

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15. Further Exam-Style Questions

Question 16

Find $\displaystyle\sum_{r=1}^{n} r(r+1)(r+2)$.

Solution

$r(r+1)(r+2) = \dfrac{1}{4}[r(r+1)(r+2)(r+3) - (r-1)r(r+1)(r+2)]$.

This telescopes: $\displaystyle\sum_{r=1}^{n} r(r+1)(r+2) = \dfrac{1}{4}n(n+1)(n+2)(n+3)$.

$\boxed{\displaystyle\sum_{r=1}^{n} r(r+1)(r+2) = \frac{n(n+1)(n+2)(n+3)}{4}}$

Question 17

Prove that $\binom{2n}{n} = \displaystyle\sum_{k=0}^{n} \binom{n}{k}^2$.

Solution

Consider choosing $n$ people from a group of $n$ men and $n$ women.

LHS: $\binom{2n}{n}$ chooses any $n$ from $2n$.

RHS: choosing $k$ men and $n-k$ women for each $k$ gives $\displaystyle\sum_{k=0}^{n} \binom{n}{k}\binom{n}{n-k} = \sum_{k=0}^{n} \binom{n}{k}^2$.

Since $\binom{n}{n-k} = \binom{n}{k}$, the identity follows. $\blacksquare$

Question 18

Prove by induction that $\displaystyle\sum_{r=1}^{n} \frac{1}{r(r+1)(r+2)} = \frac{n(n+3)}{4(n+1)(n+2)}$.

Solution

$\dfrac{1}{r(r+1)(r+2)} = \dfrac{1}{2}\!\left(\dfrac{1}{r(r+1)} - \dfrac{1}{(r+1)(r+2)}\right)$.

This telescopes: $\dfrac{1}{2}\!\left(\dfrac{1}{2} - \dfrac{1}{(n+1)(n+2)}\right) = \dfrac{(n+1)(n+2)-2}{4(n+1)(n+2)} = \dfrac{n^2+3n}{4(n+1)(n+2)} = \dfrac{n(n+3)}{4(n+1)(n+2)}$. $\blacksquare$