Sequences & Series

Q1. (4 marks)

• Definition: An AP is a sequence in which each term differs from the previous term by a fixed constant $d$ (the common difference): $a, a+d, a+2d,\dots$ (1)
• Write $S_n = a + (a+d) + \cdots + [a+(n-1)d]$ and the reverse $S_n = [a+(n-1)d] + \cdots + a$. (1)
• Adding termwise, each of the $n$ pairs sums to $2a+(n-1)d$: $2S_n = n[2a+(n-1)d]$. (1)
• Hence $S_n = \frac{n}{2}[2a+(n-1)d]$. (1) (Why: reversing exploits symmetry so pairing gives a constant sum.)

Q2. (4 marks)

• $a=18$, $r = \frac{6}{18} = \frac13$. (1)
• $|r|<1$ so the series converges. (1)
• $S_\infty = \frac{a}{1-r} = \frac{18}{1-\frac13} = \frac{18}{\frac23} = 27$. (1)
• Condition: an infinite GP converges iff $|r|<1$ (then $r^n\to 0$). (1)

Q3. (5 marks)

• $a+3d = 11$, $a+6d = 20$. (1)
• Subtract: $3d = 9 \Rightarrow d = 3$. (1)
• $a = 11 - 9 = 2$. (1)
• $S_{15} = \frac{15}{2}[2(2)+14(3)] = \frac{15}{2}[4+42] = \frac{15}{2}(46)$. (1)
• $= 15\times 23 = 345$. (1)

Q4. (4 marks)

• Definition: $a,b,c$ are in HP if their reciprocals $\frac1a,\frac1b,\frac1c$ are in AP; $b$ is the HM of $a$ and $c$. (1)
• AP condition on reciprocals: $\frac{2}{b} = \frac1a + \frac1c$. (1)
• $\frac{2}{b} = \frac{a+c}{ac}$. (1)
• $\therefore b = \frac{2ac}{a+c}$. (1)

Q5. (4 marks)

• Statement: For $a,b>0$, $\frac{a+b}{2} \ge \sqrt{ab}$, equality iff $a=b$. (1)
• Consider $(\sqrt a - \sqrt b)^2 \ge 0$. (1)
• Expand: $a - 2\sqrt{ab} + b \ge 0 \Rightarrow a+b \ge 2\sqrt{ab}$. (1)
• Divide by 2: $\frac{a+b}{2}\ge\sqrt{ab}$; equality when $\sqrt a=\sqrt b$, i.e. $a=b$. (1)

Q6. (5 marks)

• $\sum = 2\sum r^2 + 3\sum r + \sum 1$, $n=20$. (1)
• $\sum r^2 = \frac{20\cdot21\cdot41}{6} = 2870$. (1)
• $\sum r = \frac{20\cdot21}{2} = 210$; $\sum 1 = 20$. (1)
• $2(2870) + 3(210) + 20 = 5740 + 630 + 20$. (1)
• $= 6390$. (1)

Q7. (4 marks)

• Partial fractions: $\frac{1}{r(r+1)} = \frac1r - \frac{1}{r+1}$. (1)
• Sum telescopes: $\left(1-\frac12\right)+\left(\frac12-\frac13\right)+\cdots+\left(\frac1n-\frac1{n+1}\right)$. (1)
• Intermediate terms cancel, leaving $1 - \frac{1}{n+1}$. (1)
• $= \frac{n}{n+1}$. (1)

Q8. (4 marks)

• General term: $T_{k+1} = \binom{8}{k}2^{8-k}x^k$. (1)
• For $x^5$, $k=5$. (1)
• Coefficient $= \binom{8}{5}2^{3} = 56 \times 8$. (1)
• $= 448$. (1)

Q9. (3 marks)

• $(1+x)^{1/2} \approx 1 + \frac12 x - \frac18 x^2$, with $x=0.02$. (1)
• $= 1 + 0.01 - \frac18(0.0004) = 1 + 0.01 - 0.00005$. (1)
• $= 1.00995 \approx 1.0100$ (4 d.p.). (1)

Q10. (3 marks)

• Base: $n=1$: LHS $=1$, RHS $=\left(\frac{1\cdot2}{2}\right)^2 = 1$. ✓ (1)
• Inductive step: assume $\sum_{r=1}^{k} r^3 = \left(\frac{k(k+1)}{2}\right)^2$. Then $\sum_{r=1}^{k+1} r^3 = \frac{k^2(k+1)^2}{4} + (k+1)^3 = (k+1)^2\left[\frac{k^2}{4}+(k+1)\right] = (k+1)^2\cdot\frac{k^2+4k+4}{4} = \left(\frac{(k+1)(k+2)}{2}\right)^2$. (1)
• This matches the formula with $k+1$, so by induction it holds for all $n\ge1$. (1)

[
  {"claim":"Q2 infinite GP sum = 27","code":"a=18; r=Rational(1,3); result = (a/(1-r))==27"},
  {"claim":"Q3 S15 = 345","code":"a=2; d=3; n=15; S=n*(2*a+(n-1)*d)/2; result = S==345"},
  {"claim":"Q6 sum = 6390","code":"r=symbols('r'); result = summation(2*r**2+3*r+1,(r,1,20))==6390"},
  {"claim":"Q7 telescoping sum = n/(n+1)","code":"r,n=symbols('r n',positive=True,integer=True); result = simplify(summation(1/(r*(r+1)),(r,1,n)) - n/(n+1))==0"},
  {"claim":"Q8 coefficient of x^5 in (2+x)^8 is 448","code":"x=symbols('x'); result = expand((2+x)**8).coeff(x,5)==448"}
]