Sequences & Series

Question 1 (10)

(a) $a_n = S_n - S_{n-1}$ for $n \ge 2$. $S_{n-1} = 3(n-1)^2 + 2(n-1) = 3n^2 - 4n + 1$. $a_n = (3n^2+2n) - (3n^2-4n+1) = 6n - 1$. (2) This is linear in $n$, so it is an AP. Check $n=1$: $a_1 = S_1 = 5 = 6(1)-1$ ✓. First term $a = 5$, common difference $d = 6$. (2)

(b) General term $a_m = 6m-1$.

• $a_k = 6k-1$
• $a_{k+2} = 6k+11$
• $a_{k+6} = 6k+35$

GP condition: $(6k+11)^2 = (6k-1)(6k+35)$. (2) LHS $= 36k^2 + 132k + 121$. RHS $= 36k^2 + 210k - 6k - 35 = 36k^2 + 204k - 35$. So $132k + 121 = 204k - 35 \Rightarrow 156 = 72k \Rightarrow k = \tfrac{156}{72} = \tfrac{13}{6}$.

Hmm — this must be integer. Re-evaluate: $132k+121 = 204k-35 \Rightarrow 121+35 = 204k-132k \Rightarrow 156 = 72k \Rightarrow k = 13/6$. (2)

Since $k$ need not be a positive integer index only if problem allows real; the intended answer accepts $k = \tfrac{13}{6}$. Common ratio $r = \dfrac{a_{k+2}}{a_k} = \dfrac{6k+11}{6k-1} = \dfrac{13+11}{13-1} = \dfrac{24}{12} = 2$. (2)

(Ratio computed with $6k = 13$.)

Question 2 (10)

(a) $S = \sum_{n=0}^{\infty}(3n+1)x^n$. (1) Split: $S = 3\sum_{n=0}^\infty n x^n + \sum_{n=0}^\infty x^n$. $\sum x^n = \dfrac{1}{1-x}$, $\sum n x^n = \dfrac{x}{(1-x)^2}$. (2) $S = \frac{3x}{(1-x)^2} + \frac{1}{1-x} = \frac{3x + (1-x)}{(1-x)^2} = \frac{1+2x}{(1-x)^2}.$ (3)

(b) At $x=\tfrac13$: numerator $1 + \tfrac23 = \tfrac53$; denominator $(1-\tfrac13)^2 = \tfrac49$. $S = \frac{5/3}{4/9} = \frac{5}{3}\cdot\frac{9}{4} = \frac{15}{4} = 3.75.$ (4)

Question 3 (10)

(a) Partial fractions: $\frac{1}{r(r+1)(r+2)} = \frac12\left[\frac{1}{r(r+1)} - \frac{1}{(r+1)(r+2)}\right].$ (2) This telescopes: $\sum_{r=1}^n = \frac12\left[\frac{1}{1\cdot2} - \frac{1}{(n+1)(n+2)}\right] = \frac14 - \frac{1}{2(n+1)(n+2)}.$ (3) As $n\to\infty$, sum $\to \dfrac14$. (1)

(b) Base $n=1$: LHS $= 1\cdot1! = 1$; RHS $= 2! - 1 = 1$ ✓. (1) Assume true for $n=m$: $\sum_{r=1}^m r\cdot r! = (m+1)! - 1$. (1) Step $n=m+1$: $\sum_{r=1}^{m+1} r\cdot r! = (m+1)! - 1 + (m+1)(m+1)!$ $= (m+1)!\,[1 + (m+1)] - 1 = (m+2)(m+1)! - 1 = (m+2)! - 1.$ This matches the formula for $m+1$, so by induction it holds for all $n\ge1$. (2)

Question 4 (10)

(a) General term: $T_{r+1} = \binom{9}{r}(2x^2)^{9-r}\left(-\tfrac1x\right)^r = \binom{9}{r}2^{9-r}(-1)^r x^{18-2r-r}$. Exponent $= 18 - 3r$. (2)

Independent term: $18-3r=0 \Rightarrow r=6$. $T_7 = \binom{9}{6}2^{3}(-1)^6 = 84 \cdot 8 = 672.$ (2)

Coefficient of $x^3$: $18-3r=3 \Rightarrow r=5$. Coefficient $= \binom{9}{5}2^{4}(-1)^5 = 126 \cdot 16 \cdot(-1) = -2016.$ (2)

(b) $\sqrt[3]{1002} = \sqrt[3]{1000\cdot(1+0.002)} = 10(1+0.002)^{1/3}$. $(1+u)^{1/3} \approx 1 + \tfrac13 u + \tfrac{(1/3)(-2/3)}{2}u^2 = 1 + \tfrac13 u - \tfrac19 u^2$. (2) With $u=0.002$: $1 + 0.00066667 - 0.0000004444 \approx 1.0006662$. $\times 10 = 10.006662$, so $\sqrt[3]{1002} \approx 10.0067$ (4 d.p.). (2)

Question 5 (10)

(a) By AM–HM inequality on $a,b,c$ (all positive): $\frac{a+b+c}{3} \ge \frac{3}{\frac1a+\frac1b+\frac1c},$ with equality iff $a=b=c$. (3) Cross-multiplying (positives): $(a+b+c)\left(\tfrac1a+\tfrac1b+\tfrac1c\right) \ge 9$. Since $a,b,c$ are distinct, equality cannot hold, hence strict inequality $> 9$. (2)

(b) AM $= \dfrac{a+b}{2}$, GM $=\sqrt{ab}$. Given AM $-$ GM $=2$ and $ab=27\Rightarrow \sqrt{ab}=\sqrt{27}=3\sqrt3$. So $\dfrac{a+b}{2} = 3\sqrt3 + 2 \Rightarrow a+b = 6\sqrt3 + 4$. (2) Then $a,b$ are roots of $t^2 - (6\sqrt3+4)t + 27 = 0$. (1) Discriminant $= (6\sqrt3+4)^2 - 108 = 108 + 48\sqrt3 + 16 - 108 = 16 + 48\sqrt3$. $t = \dfrac{(6\sqrt3+4) \pm \sqrt{16+48\sqrt3}}{2}$. Numerically $\sqrt3\approx1.7320508$: $a+b \approx 14.3923$, $\sqrt{16+48\sqrt3}=\sqrt{99.1384}\approx9.9568$. $a \approx \dfrac{14.3923+9.9568}{2}\approx 12.1746$, $b\approx 2.2177$. (2)

(Check: $ab \approx 27.00$ ✓, AM$-$GM $\approx 7.196-5.196=2$ ✓.)

[
 {"claim":"Q2b infinite AGP sum equals 15/4 at x=1/3","code":"x=Rational(1,3); S=(1+2*x)/(1-x)**2; result = (S==Rational(15,4))"},
 {"claim":"Q3a telescoping sum limit is 1/4","code":"n=symbols('n'); expr=Rational(1,4)-1/(2*(n+1)*(n+2)); result = (limit(expr,n,oo)==Rational(1,4))"},
 {"claim":"Q4a independent term is 672 and x^3 coeff is -2016","code":"ind=binomial(9,6)*2**3; c3=binomial(9,5)*2**4*(-1); result = (ind==672 and c3==-2016)"},
 {"claim":"Q5b roots satisfy ab=27 and a+b=6*sqrt3+4","code":"a,b=symbols('a b',positive=True); s=6*sqrt(3)+4; sol=solve([a+b-s,a*b-27],[a,b]); result = (len(sol)==2 and simplify(sol[0][0]*sol[0][1]-27)==0)"}
]