1. Tardigrade
  2. Question
  3. Mathematics
  4. Let an n=0∞ be a sequence such that a 0= a 1=0 and a n +2=2 a n +1- a n +1 for all n ≥ 0. Then, displaystyle∑ n =2∞ ( a n /7 n ) is equal to

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Q. Let $\left\{a_{n}\right\}_{n=0}^{\infty}$ be a sequence such that $a _{0}= a _{1}=0$ and $a _{ n +2}=2 a _{ n +1}- a _{ n }+1$ for all $n \geq 0$. Then, $\displaystyle\sum_{ n =2}^{\infty} \frac{ a _{ n }}{7^{ n }}$ is equal to

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Solution:

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