Question

Let ${\left\{a_n\right\}}_{n=0}^{\infty }$ be a sequence such that $a_0=a_1=0$ and $a_{n+2}=2a_{n+1}-a_n+1$ for all $n\ge 0$. Then, $\sum _{n=2}^{\infty }\frac{a_n}{7^n}$ is equal to

• A. $\frac{6}{343}$
• B. $\frac{7}{216}$
• C. $\frac{8}{343}$
• D. $\frac{49}{216}$

Answer 1

Answer: (B)

$a_2=1,a_3=3a_4=6$
$a_n=\frac{n(n-1)}{2}$
$S=\sum _{n=2}^{\infty }\frac{n(n-1)}{2\left(7^n\right)}$
$S=\frac{1}{7^2}+\frac{3}{7^3}+\frac{6}{7^4}+\frac{10}{7^5}+\frac{15}{7^5}+\dots \dots$
$\frac{S}{7}=\frac{1}{7^3}+\frac{3}{7^4}+\frac{6}{7^5}+\frac{10}{7^6}+\dots$
$6\frac{S}{7}=\frac{1}{7^2}+\frac{2}{7^3}+\frac{3}{7^4}+\frac{4}{7^5}+\dots$
$6\frac{S}{7^2}=\frac{1}{7^3}+\frac{2}{7^4}+\frac{3}{7^5}+\dots$
$6\frac{S}{7}\cdot \frac{6}{7}=\frac{1}{7^2}+\frac{1}{7^3}+\dots =\frac{1/7^2}{1-1/7}$
$6\times 6\frac{S}{7^2}=\cdot \frac{1}{7\times 6}$
$S=\frac{7}{6^3}=\frac{7}{216}$
Alternate
$a_{n+2}=2a_{n+1}-a_n+1$
$\Rightarrow \frac{a_{n+2}}{7^{n+2}}=\frac{2}{7}\frac{a_{n+1}}{7^{n+1}}-\frac{1}{49}\frac{a_n}{7^n}+\frac{1}{7^{n+2}}$
$\Rightarrow \sum _{n=2}^{\infty }\frac{a_{n+2}}{7^{n+2}}=\frac{2}{7}\sum _{n=2}^{\infty }\frac{a_{n+1}}{7^{n+1}}-\frac{1}{49}\sum _{n=2}^{\infty }\frac{a_n}{7^n}+\sum _{n=2}^{\infty }\frac{1}{7^{n+2}}$
Let $\sum _{n=2}^{\infty }\frac{a_n}{7^n}=p$
$\Rightarrow \left(p-\frac{a_2}{7^2}-\frac{a_3}{7^3}\right)=\frac{2}{7}\left(p-\frac{a_2}{7^2}\right)-\frac{1}{49}p+\frac{1/7^4}{1-\frac{1}{7}}$
$\because a_2=1,a_3=3$
$\Rightarrow p-\frac{1}{49}-\frac{3}{343}=\frac{2}{7}p-\frac{2}{7^3}-\frac{p}{49}+\frac{1}{6.7^3}$
$\Rightarrow p=\frac{7}{216}$