Level 3 — ProductionSequences & Series

Sequences & Series

⏱ 45 minutes60 marksprintable — key stays hidden on paper

Level: 3 (From-scratch derivations, code-from-memory, explain-out-loud) Time limit: 45 minutes Total marks: 60

Instructions: Show every step. Where a derivation is requested, start from first principles — do not merely quote a result. "Explain-out-loud" parts require reasoning in words alongside algebra.

Q1. (10 marks) — Derive the sum of an infinite GP + convergence proof.

(a) Starting from the finite sum $S_n = a + ar + \dots + ar^{n-1}$ , derive a closed form for $S_n$ (state the method). (4)

(b) Explain out loud why the infinite sum converges only when $|r| < 1$ , and prove that in that case $S_\infty = \dfrac{a}{1-r}$ . (4)

Q2. (10 marks) — Prove the AM–GM–HM inequality chain for two positives.

For positive reals $a, b$ , prove from scratch that $\frac{a+b}{2} \ge \sqrt{ab} \ge \frac{2ab}{a+b},$ stating the equality condition. Explain out loud where each inequality "comes from." (8) Hence show that for $a=4,\ b=9$ the three means are $6.5,\ 6,\ \tfrac{72}{13}$ and confirm the ordering. (2)

Q3. (12 marks) — Σ-formulae and telescoping.

(a) Prove by first principles (using the algebraic identity for $(k+1)^3 - k^3$ ) that $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}.$ (6)

(b) Using telescoping, evaluate $\sum_{k=1}^{n} \frac{1}{k(k+1)}$ and state its limit as $n \to \infty$ . (4)

Q4. (10 marks) — Mathematical induction.

Prove by induction that for all integers $n \ge 1$ : $\sum_{k=1}^{n} k \cdot 2^{k} = (n-1)2^{n+1} + 2.$ Clearly label base case, inductive hypothesis, and inductive step, and explain out loud what the hypothesis is being used for. (10)

Q5. (10 marks) — Binomial general term.

(a) Write the general term $T_{r+1}$ of $\left(2x^2 - \dfrac{3}{x}\right)^{9}$ . (3)

(b) Find the term independent of $x$ . (4)

(c) Using the binomial series for rational index, find an approximation of $\sqrt{1.02}$ to 4 decimal places (keep terms up to the $x^2$ term). (3)

Q6. (8 marks) — Arithmetic–geometric series (from-scratch sum).

Derive, using the "multiply by common ratio and subtract" technique, a closed form for $S = \sum_{k=1}^{n} k\, r^{k-1}, \quad r \ne 1.$ Show all shifting/subtraction steps, then state $\displaystyle \sum_{k=1}^{\infty} k r^{k-1}$ for $|r|<1$ . (8)

Answer keyMark scheme & solutions

Q1 (10)

(a) Method: multiply-and-subtract. $S_n = a + ar + \dots + ar^{n-1}$ , $rS_n = ar + \dots + ar^{n}$ . Subtract: $S_n - rS_n = a - ar^n \Rightarrow S_n(1-r) = a(1-r^n)$ . $S_n = \frac{a(1-r^n)}{1-r}, \quad r \ne 1.$ (setup 1, subtraction 2, closed form 1)

(b) As $n\to\infty$ , $r^n\to 0$ iff $|r|<1$ ; for $|r|\ge 1$ the term $ar^n$ does not vanish (diverges or oscillates), so no finite limit. When $|r|<1$ : $S_\infty = \lim_{n\to\infty}\frac{a(1-r^n)}{1-r} = \frac{a(1-0)}{1-r} = \frac{a}{1-r}.$ (convergence condition explained 2, limit + result 2)

(c) $a = \frac34,\ r=\frac14$ : $S = \dfrac{3/4}{1-1/4} = \dfrac{3/4}{3/4} = 1.$ (2)

Q2 (10)

AM ≥ GM: $(\sqrt a - \sqrt b)^2 \ge 0 \Rightarrow a + b - 2\sqrt{ab} \ge 0 \Rightarrow \frac{a+b}{2} \ge \sqrt{ab}$ . Equality iff $a=b$ . (3)

GM ≥ HM: From AM≥GM applied to $\frac1a,\frac1b$ : $\frac{1/a+1/b}{2}\ge\frac{1}{\sqrt{ab}}$ , i.e. $\frac{a+b}{2ab}\ge\frac1{\sqrt{ab}}$ , invert (both positive): $\frac{2ab}{a+b}\le\sqrt{ab}$ . Equality iff $a=b$ . (4)

Explain: each inequality springs from a perfect-square being non-negative (directly, or applied to reciprocals). (1)

Numeric: $a=4,b=9$ : AM $=\frac{13}{2}=6.5$ ; GM $=\sqrt{36}=6$ ; HM $=\frac{2\cdot36}{13}=\frac{72}{13}\approx5.538$ . Ordering $6.5>6>5.538$ . ✓ (2)

Q3 (12)

(a) Identity: $(k+1)^3 - k^3 = 3k^2 + 3k + 1$ . Sum $k=1..n$ (telescoping LHS): $(n+1)^3 - 1 = 3\sum k^2 + 3\sum k + n$ . $\sum k = \frac{n(n+1)}{2}$ , so $3\sum k^2 = (n+1)^3 - 1 - \frac{3n(n+1)}{2} - n$ . Simplify: $(n+1)^3 - 1 - n = n^3+3n^2+2n = n(n+1)(n+2)$ ... expand carefully: $3\sum k^2 = n^3+3n^2+3n+1-1-n - \frac{3n(n+1)}{2}= n^3+3n^2+2n-\frac{3n^2+3n}{2}$ $= \frac{2n^3+6n^2+4n-3n^2-3n}{2}=\frac{2n^3+3n^2+n}{2}=\frac{n(n+1)(2n+1)}{2}$ . $\sum k^2 = \frac{n(n+1)(2n+1)}{6}.$ (identity 1, telescope 2, algebra 3)

(b) $\frac{1}{k(k+1)}=\frac1k-\frac1{k+1}$ . Telescoping sum $=1-\frac1{n+1}=\frac{n}{n+1}$ . Limit $=1$ . (4)

(c) $n=20$ : $\frac{20\cdot21\cdot41}{6}=\frac{17220}{6}=2870.$ (2)

Q4 (10)

Let $P(n): \sum_{k=1}^n k2^k = (n-1)2^{n+1}+2.$

Base $n=1$ : LHS $=1\cdot2=2$ ; RHS $=(0)2^2+2=2$ . ✓ (2)

Hypothesis: assume $P(m)$ true: $\sum_{k=1}^m k2^k=(m-1)2^{m+1}+2$ . (2)

Step: $\sum_{k=1}^{m+1}k2^k = (m-1)2^{m+1}+2 + (m+1)2^{m+1}$ $= 2^{m+1}[(m-1)+(m+1)]+2 = 2^{m+1}(2m)+2 = m\cdot2^{m+2}+2$ $=((m+1)-1)2^{(m+1)+1}+2.$ This is $P(m+1)$ . ✓ (hypothesis used to replace the sum of first $m$ terms — 4)

By induction, true for all $n\ge1$ . (2)

Q5 (10)

(a) $T_{r+1}=\binom{9}{r}(2x^2)^{9-r}\left(-\frac3x\right)^r = \binom9r 2^{9-r}(-3)^r x^{18-2r-r}=\binom9r2^{9-r}(-3)^r x^{18-3r}.$ (3)

(b) Independent: $18-3r=0\Rightarrow r=6$ . $T_7=\binom96 2^{3}(-3)^6 = 84\cdot8\cdot729 = 489888.$ (4)

(c) $\sqrt{1.02}=(1+0.02)^{1/2}\approx 1+\frac12(0.02)+\frac{\frac12(-\frac12)}{2}(0.02)^2$ $=1+0.01-0.00005=1.00995\approx 1.0100.$ (3)

Q6 (8)

$S = \sum_{k=1}^n k r^{k-1} = 1 + 2r + 3r^2+\dots+nr^{n-1}.$ $rS = r + 2r^2+\dots+(n-1)r^{n-1}+nr^n.$ Subtract: $S - rS = 1 + r + r^2+\dots+r^{n-1} - nr^n = \frac{1-r^n}{1-r}-nr^n.$ $S = \frac{1-r^n}{(1-r)^2} - \frac{nr^n}{1-r}.$ For $|r|<1$ , $r^n\to0,\ nr^n\to0$ : $\displaystyle\sum_{k=1}^\infty kr^{k-1}=\frac{1}{(1-r)^2}.$ (setup 2, shift 2, subtraction/geometric 2, closed form 1, infinite case 1)

[
{"claim":"Q1c infinite GP sum = 1","code":"a=Rational(3,4); r=Rational(1,4); result = (a/(1-r))==1"},
{"claim":"Q3c sum k^2 to 20 = 2870","code":"result = sum(k**2 for k in range(1,21))==2870"},
{"claim":"Q5b independent term = 489888","code":"from sympy import binomial; result = binomial(9,6)*2**3*(-3)**6==489888"},
{"claim":"Q4 identity holds for n=1..6","code":"result = all(sum(k*2**k for k in range(1,n+1))==(n-1)*2**(n+1)+2 for n in range(1,7))"},
{"claim":"Q6 closed form matches direct sum for r=1/2,n=5","code":"r=Rational(1,2); n=5; lhs=sum(k*r**(k-1) for k in range(1,n+1)); rhs=(1-r**n)/(1-r)**2-n*r**n/(1-r); result = simplify(lhs-rhs)==0"}
]