Level 5 — MasterySequences & Series

Sequences & Series

⏱ 75 minutes50 marksprintable — key stays hidden on paper

Difficulty: Level 5 — Mastery (cross-domain: math + physics + coding; build & prove) Time limit: 75 minutes Total marks: 50

Answer all questions. Full derivations and proofs are required. Calculators are not permitted; leave answers exact unless an approximation is requested.

Question 1 — Convergence, Physics & Telescoping (18 marks)

A bouncing ball is dropped from height $h_0 = 2\text{ m}$ . After each bounce it rebounds to a fraction $r$ ( $0 < r < 1$ ) of its previous peak height. Use $g = 10\ \text{m s}^{-2}$ throughout.

(a) From first principles, prove that the infinite geometric series $\sum_{n=0}^{\infty} a r^n$ converges iff $|r| < 1$ , and derive its sum. (5)

(b) The total vertical distance travelled by the ball before coming to rest is $D = h_0 + 2\sum_{n=1}^{\infty} h_0 r^n.$ Explain physically the factor $2$ and the missing $2$ on the first term, then show that $D = h_0\,\frac{1+r}{1-r}.$ Evaluate $D$ for $r = \tfrac12$ . (4)

(c) The time for a free flight of peak height $h$ (up and down) is $t = 2\sqrt{2h/g}$ . Show that the total time until rest is finite and equals $T = \sqrt{\frac{2h_0}{g}}\left(\frac{1+\sqrt{r}}{1-\sqrt{r}}\right).$ Evaluate $T$ for $r=\tfrac12$ , $h_0 = 2$ , giving your answer in exact surd form. (5)

(d) Using telescoping, evaluate in closed form $S = \sum_{k=1}^{N} \frac{1}{h_0 r^{k-1}\,+\,\text{(a distractor)}}\quad\Longrightarrow\quad \text{evaluate instead } \sum_{k=1}^{N}\frac{1}{k(k+1)},$ and hence state $\lim_{N\to\infty} \sum_{k=1}^{N}\frac{1}{k(k+1)}$ . (4)

Question 2 — Induction, Binomial & a Coding Algorithm (18 marks)

(a) State the Binomial Theorem for a positive integer index $n$ and prove it by induction on $n$ , clearly using Pascal's rule $\binom{n}{k-1}+\binom{n}{k}=\binom{n+1}{k}$ . (6)

(b) Consider the recurrence used by a program to build Pascal's triangle:

C[0] = 1
for k = 1 .. n:
    C[k] = C[k-1] * (n - k + 1) / k

Prove by induction on $k$ that after the loop C[k] $=\binom{n}{k}$ for all $0\le k\le n$ . (4)

(c) Find the term independent of $x$ in the expansion of $\left(2x^2 - \frac{1}{x}\right)^{9}.$ State the general term first. (4)

(d) Using the binomial series for a rational index, estimate $\sqrt[3]{1002}$ correct to 4 decimal places, showing the terms you keep and bounding the truncation error. (4)

Question 3 — Means, Inequalities & Strong Induction (14 marks)

(a) For positive reals $a,b$ define AM, GM, HM. Prove $\text{AM}\ge\text{GM}\ge\text{HM}$ , with equality iff $a=b$ . Deduce that $\text{GM}^2 = \text{AM}\cdot\text{HM}$ . (6)

(b) The numbers $x, y, z$ are in AP, GP and HP respectively (three separate progressions sharing the same two end terms $a=4$ , $c=9$ as first/third terms). Compute the middle terms and verify GM $^2$ = AM·HM numerically. (4)

(c) Every integer $n\ge 2$ can be written as a product of primes. Prove this using strong induction, and explain precisely where the strong hypothesis (not ordinary induction) is essential. (4)

Answer keyMark scheme & solutions

Question 1

(a) (5) Partial sum $S_N = a + ar + \dots + ar^{N-1}$ . Multiply by $r$ : $rS_N = ar+\dots+ar^N$ . Subtract: $S_N(1-r) = a(1-r^N)$ , so for $r\ne1$ , $S_N = a\dfrac{1-r^N}{1-r}$ . (2) As $N\to\infty$ : $r^N\to 0$ iff $|r|<1$ ; if $|r|\ge1$ the term $ar^{N-1}\not\to0$ so the series diverges. (2) Hence for $|r|<1$ , $S = \dfrac{a}{1-r}$ . (1)

(b) (4) The ball first falls $h_0$ (one trip). Every subsequent peak $h_0 r^n$ is reached up and down, hence factor $2$ ; the first drop is counted once (no matching upward pre-drop), so it lacks the factor $2$ . (2) $D = h_0 + 2h_0\dfrac{r}{1-r} = h_0\dfrac{(1-r)+2r}{1-r} = h_0\dfrac{1+r}{1-r}.$ (1) $r=\tfrac12$ : $D = 2\cdot\dfrac{3/2}{1/2} = 2\cdot3 = 6\text{ m}.$ (1)

(c) (5) Initial fall time $= \sqrt{2h_0/g}$ . Bounce $n$ has peak $h_0 r^n$ , flight time $2\sqrt{2h_0 r^n/g} = 2\sqrt{2h_0/g}(\sqrt r)^n$ . (2) $T = \sqrt{\tfrac{2h_0}{g}} + 2\sqrt{\tfrac{2h_0}{g}}\sum_{n=1}^\infty (\sqrt r)^n = \sqrt{\tfrac{2h_0}{g}}\Big(1 + \tfrac{2\sqrt r}{1-\sqrt r}\Big) = \sqrt{\tfrac{2h_0}{g}}\dfrac{1+\sqrt r}{1-\sqrt r}.$ Finite since $\sqrt r<1$ . (2) $r=\tfrac12,h_0=2,g=10$ : $\sqrt{2\cdot2/10}=\sqrt{0.4}=\tfrac{\sqrt{10}}{5}$ . Factor $\dfrac{1+\tfrac1{\sqrt2}}{1-\tfrac1{\sqrt2}} = \dfrac{\sqrt2+1}{\sqrt2-1} = (\sqrt2+1)^2 = 3+2\sqrt2.$ $T = \dfrac{\sqrt{10}}{5}(3+2\sqrt2)\text{ s}.$ (1)

(d) (4) $\dfrac1{k(k+1)} = \dfrac1k - \dfrac1{k+1}$ (partial fractions). (1) Telescoping: $\sum_{k=1}^N = 1 - \dfrac1{N+1} = \dfrac{N}{N+1}.$ (2) Limit $= 1$ . (1)

Question 2

(a) (6) Statement: $(x+y)^n = \sum_{k=0}^n \binom{n}{k}x^{n-k}y^k$ . (1) Base $n=1$ : RHS $= x + y$ . ✓ (1) Assume true for $n$ . Then $(x+y)^{n+1} = (x+y)\sum_k\binom nk x^{n-k}y^k$ . (1) Expand: $\sum_k\binom nk x^{n-k+1}y^k + \sum_k\binom nk x^{n-k}y^{k+1}$ ; reindex second sum ( $k\to k-1$ ). (1) Coefficient of $x^{n+1-k}y^k$ : $\binom nk + \binom n{k-1} = \binom{n+1}{k}$ by Pascal's rule. (1) So $(x+y)^{n+1}=\sum_k\binom{n+1}{k}x^{n+1-k}y^k$ ; by induction proved for all $n\ge1$ . (1)

(b) (4) Claim $C[k]=\binom nk$ . Base $k=0$ : $C[0]=1=\binom n0$ . ✓ (1) Assume $C[k-1]=\binom n{k-1}$ . Loop sets $C[k]=\binom n{k-1}\cdot\dfrac{n-k+1}{k}$ . (1) $\binom n{k-1}\dfrac{n-k+1}{k} = \dfrac{n!}{(k-1)!(n-k+1)!}\cdot\dfrac{n-k+1}{k} = \dfrac{n!}{k!(n-k)!}=\binom nk.$ (1) By induction holds for all $0\le k\le n$ . (1)

(c) (4) General term $T_{r+1}=\binom9r (2x^2)^{9-r}(-x^{-1})^r = \binom9r 2^{9-r}(-1)^r x^{18-2r-r}.$ (2) Power of $x$ : $18-3r=0\Rightarrow r=6$ . (1) $T_7 = \binom96 2^{3}(-1)^6 = 84\cdot8 = 672.$ (1)

(d) (4) $\sqrt[3]{1002}=10\sqrt[3]{1.002}=10(1+0.002)^{1/3}.$ (1) $(1+u)^{1/3}=1+\tfrac13 u-\tfrac19 u^2+\dots$ , $u=0.002$ : $\tfrac13u=6.6\overline6\times10^{-4}$ , $\tfrac19u^2\approx4.4\times10^{-7}$ (negligible at 4 dp). (1) $\sqrt[3]{1.002}\approx1.0006667-0.0000004=1.0006663$ . (1) $\sqrt[3]{1002}\approx10.0066633 \Rightarrow \boxed{10.0067}$ (4 dp). Truncation error $<10\cdot\tfrac{5}{81}u^3\approx10^{-8}$ . (1)

Question 3

(a) (6) Definitions: $\text{AM}=\tfrac{a+b}2$ , $\text{GM}=\sqrt{ab}$ , $\text{HM}=\dfrac{2ab}{a+b}$ . (1) AM≥GM: $(\sqrt a-\sqrt b)^2\ge0\Rightarrow a+b\ge2\sqrt{ab}$ ; equality iff $a=b$ . (2) GM≥HM: from AM≥GM applied form, $\dfrac{2ab}{a+b}\le\sqrt{ab}$ ⇔ $2\sqrt{ab}\le a+b$ (same inequality, since $\sqrt{ab}>0$ ). Equality iff $a=b$ . (2) $\text{AM}\cdot\text{HM}=\tfrac{a+b}2\cdot\dfrac{2ab}{a+b}=ab=\text{GM}^2$ . (1)

(b) (4) With end terms $a=4,c=9$ : AP middle $=\tfrac{4+9}2=6.5$ ; GP middle $=\sqrt{36}=6$ ; HP middle $=\dfrac{2\cdot4\cdot9}{13}=\dfrac{72}{13}\approx5.538$ . (2) Check GM $^2=36$ ; AM·HM $=6.5\cdot\tfrac{72}{13}=\tfrac{468}{13}=36$ . ✓ (2)

(c) (4) Base $n=2$ : prime, product of one prime. (1) Strong step: assume every $m$ with $2\le m<n$ is a product of primes. If $n$ prime, done. Else $n=ab$ with $2\le a,b<n$ . (1) By strong hypothesis $a,b$ are products of primes, so $n=ab$ is too. (1) Ordinary induction (only $n-1$ ) fails because factors $a,b$ are generally not $n-1$ ; we need the hypothesis for all smaller values. (1)

[
  {"claim":"Total distance D=6 for h0=2,r=1/2","code":"h0,r=2,Rational(1,2); D=h0*(1+r)/(1-r); result=(D==6)"},
  {"claim":"Time factor (1+1/sqrt2)/(1-1/sqrt2)=3+2sqrt2","code":"expr=(1+1/sqrt(2))/(1-1/sqrt(2)); result=simplify(expr-(3+2*sqrt(2)))==0"},
  {"claim":"Telescoping sum to N equals N/(N+1)","code":"N=symbols('N',positive=True,integer=True); k=symbols('k'); s=summation(1/(k*(k+1)),(k,1,N)); result=simplify(s-N/(N+1))==0"},
  {"claim":"Independent term of (2x^2-1/x)^9 is 672","code":"x=symbols('x'); from sympy import expand; e=expand((2*x**2-1/x)**9); result=(e.coeff(x,0)==672)"},
  {"claim":"AM*HM=GM^2 for a=4,c=9 gives 36","code":"a,c=4,9; AM=Rational(a+c,2); HM=Rational(2*a*c,a+c); result=(AM*HM==36)"}
]