Alternating terms, even terms and products · source reference

Printed question

A geometric series G is given by $e-e^{1+\alpha}+e^{1+2\alpha}-e^{1+3\alpha}+\cdots$, where $\alpha$ is a negative real number.

(i) Show that G is convergent. [2]

(ii) Its sum to infinity is $S$. Given that the sum to infinity of the even-numbered terms (the second, fourth, sixth, … terms) is $-\frac13S$, find $\alpha$ exactly. [4]

(iii) In another series H, each term is the modulus of the corresponding G term. Given $\alpha=-0.03$, find the greatest $n$ such that the product of H’s first $n$ terms exceeds 15000. [4]

Printed working

(i) Common ratio $r=-e^{\alpha}$; $|-e^{\alpha}|=e^{\alpha}<1$ since $\alpha<0$, so G converges.

(ii) $S=e/(1+e^{\alpha})$. Even-term sum:

$$\frac{-e^{1+\alpha}}{1-(-e^{\alpha})^2}=\frac{-e^{1+\alpha}}{1-e^{2\alpha}}=-\frac13\frac e{1+e^{\alpha}}.$$ $$\frac{e^{\alpha}}{1-e^{\alpha}}=\frac13\Rightarrow4e^{\alpha}=1\Rightarrow e^{\alpha}=\frac14\Rightarrow\alpha=-\ln4.$$

(iii) H is $e+e^{1+\alpha}+e^{1+2\alpha}+\cdots$. Product of $n$ terms:

$$e\,e^{1+\alpha}\cdots e^{1+(n-1)\alpha}=e^{1+(1+\alpha)+(1+2\alpha)+\cdots+[1+(n-1)\alpha]}=e^{\frac n2[2+(n-1)\alpha]}=e^{\frac n2(2.03-0.03n)}.$$

Given this $>15000$,

$$\frac n2(2.03-0.03n)>\ln15000\Rightarrow0.03n^2-2.03n+2\ln15000<0.$$

Roots 11.39 and 56.275. Thus $11.39<n<56.275$, greatest $n=56$.

Alternative: graph $y=e^{n(2.03-0.03n)/2}$ against $y=15000$. Intersections 11.39,56.275 yield the same interval and answer.

Teacher notes

G converges when $|r|<1$.

Even-indexed terms $T_2,T_4,T_6,\ldots=-e^{1+\alpha},-e^{1+3\alpha},-e^{1+5\alpha},\ldots$ form a GP with first term $-e^{1+\alpha}$ and common ratio $e^{2\alpha}$.

H takes the modulus of each G term, giving $e+e^{1+\alpha}+e^{1+2\alpha}+\cdots$.

Source diagrams redrawn

Printed answer key

(ii) $-\ln4$; (iii) 56

Mathematical review and qualifications

(i) Ratio is $-e^\alpha$, with modulus below one, so $G$ converges.

(ii) $S=e/(1+t)$ and the even-indexed sum is $-et/(1-t^2)$. Equality gives $t/(1-t)=1/3$, hence $t=1/4$, $\alpha=-\ln4$.

(iii) The product is $\exp[n+\alpha n(n-1)/2]$. At $\alpha=-0.03$ its logarithm is $1.015n-0.015n^2$. Solve $1.015n-0.015n^2>\ln15000$, giving approximately $11.39<n<56.275$. The greatest integer is 56; the logarithms at 56 and 57 are 9.8 and 9.12, bracketing $\ln15000\approx9.616$.

Provenance

2025 TJC H2 Math (9758) JC2 MSM: ARITHMETIC & GEOMETRIC PROGRESSIONS. [2009/IJC/Prelim/I/12] Question PDF pages 2; solution PDF pages 10, 11. The original files remain private QA records outside this native content.