Let ann=0∞ be a sequence such that a0=a1=0 and an+2=2an+1−an+1 for all n⩾0. Then, ∑n=2∞ an7n is equal to :

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

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$\frac{8}{343}$

$\frac{49}{216}$

$\frac{6}{343}$

$\frac{7}{216}$

answer is B.

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Detailed Solution

$\begin{array}{l} a_{1} = 1, a_{3} = 3 a_{4} = 6 \\ a_{n} = \frac{n (n - 1)}{2} \\ S = \sum_{n = 2}^{\infty} \frac{n (n - 1)}{2 (7^{n})} \\ S = \frac{1}{7^{2}} + \frac{3}{7^{3}} + \frac{6}{7^{4}} + \frac{10}{7^{5}} + \frac{15}{7^{5}} + \dots \dots \\ \frac{S}{7} = \frac{1}{7^{3}} + \frac{3}{7^{4}} + \frac{6}{7^{5}} + \frac{10}{7^{6}} + \dots \\ 6 \frac{S}{7} = \frac{1}{7^{2}} + \frac{2}{7^{3}} + \frac{3}{7^{4}} + \frac{4}{7^{5}} + \dots \\ 6 \frac{S}{7^{2}} = \frac{1}{7^{3}} + \frac{2}{7^{4}} + \frac{3}{7^{5}} + \dots \\ 6 \frac{S}{7} \cdot \frac{6}{7} = \frac{1}{7^{2}} + \frac{1}{7^{3}} + \dots = \frac{1 / 7^{2}}{1 - 1 / 7} \\ 6 \times 6 \frac{S}{7^{2}} = \frac{1}{7 \times 6} \\ S = \frac{7}{6^{3}} = \frac{7}{216} \end{array}$