Shift an exponential summation identity · source reference

Printed question

Given $\sum_{n=1}^N\frac{3^n}{n+1}(1-\frac1{2n})=\frac12(\frac{3^{N+1}}{N+1}-3)$, deduce $\sum_{n=1}^N\frac{3^n(2n+1)}{(n+1)(n+2)}$, leaving the answer as $a(\frac{3^{N+2}}{N+2}-b)$ where $a,b$ are constants to be determined. [3]

Printed working

$$\sum_{n=1}^N\frac{3^n}{n+1}(1-\frac1{2n})=\sum_{n=1}^N\frac{3^n(2n-1)}{2n(n+1)}.$$

Replace $n$ by $n-1$ in the requested sum:

$$\begin{aligned}\sum_{n=1}^N\frac{3^n(2n+1)}{(n+1)(n+2)}&=\sum_{n=2}^{N+1}\frac{3^{n-1}(2n-1)}{n(n+1)}\\&=\frac23\sum_{n=2}^{N+1}\frac{3^n(2n-1)}{2n(n+1)}\\&=\frac23\left[\sum_{n=1}^{N+1}\frac{3^n(2n-1)}{2n(n+1)}-\frac{3(2-1)}{2\times2}\right]\\&=\frac23\left[\frac12\left(\frac{3^{N+2}}{N+2}-3\right)-\frac34\right]\\&=\frac13\left(\frac{3^{N+2}}{N+2}-\frac92\right).\end{aligned}$$

Hence $a=\frac13,b=\frac92$.

Teacher notes

To use the given result, rewrite $f(n)$ to match its summand. $\sum_{n=2}^{N+1}T_n=\sum_{n=1}^{N+1}T_n-T_1$, where $T_n=\frac{3^n(2n-1)}{2n(n+1)}$.

Printed answer key

$a=1/3,b=9/2$

Mathematical review and qualifications

The target summand equals $(2/3)T_{n+1}$. Therefore its sum is $(2/3)(S_{N+1}-T_1)=(2/3)[\frac12(\frac{3^{N+2}}{N+2}-3)-3/4]=\frac13(\frac{3^{N+2}}{N+2}-9/2)$. Thus $a=1/3,b=9/2$.

The bases are powers 3^n, not the product 3n; extracted text loses superscripts.

Provenance

2025 TJC H2 Math (9758) JC2 MSM: SIGMA NOTATION. [modified 2020/ACJC/Prelim/I/5] Question PDF pages 2; solution PDF pages 8, 9. The original files remain private QA records outside this native content.