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(B) Given the recurrence relation $a_{n+2}-2a_{n+1}+a_{n}=1$ with $a_{0}=0, a_{1}=0$.Calculating the first few terms: $a_{2}=2(0)-0+1=1$,$a_{3}=2(1)-0

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$\frac{7}{216}$

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(B) Given the recurrence relation $a_{n+2}-2a_{n+1}+a_{n}=1$ with $a_{0}=0, a_{1}=0$.Calculating the first few terms: $a_{2}=2(0)-0+1=1$,$a_{3}=2(1)-0+1=3$,$a_{4}=2(3)-1+1=6$.The general term is $a_{n}=\frac{n(n-1)}{2}$.Let $S=\sum\limits_{n=2}^{\infty} \frac{n(n-1)}{2 \cdot 7^{n}}$.This is an arithmetico-geometric series. Using the method of generating functions or series manipulation:$S = \frac{1}{7^{2}} + \frac{3}{7^{3}} + \frac{6}{7^{4}} + \frac{10}{7^{5}} + \dots$$\frac{S}{7} = \frac{1}{7^{3}} + \frac{3}{7^{4}} + \frac{6}{7^{5}} + \dots$Subtracting these: $S(1-\frac{1}{7}) = \frac{1}{7^{2}} + \frac{2}{7^{3}} + \frac{3}{7^{4}} + \dots$$\frac{6}{7}S = \frac{1}{7^{2}} (1 + \frac{2}{7} + \frac{3}{7^{2}} + \dots) = \frac{1}{49} \cdot \frac{1}{(1-\frac{1}{7})^{2}} = \frac{1}{49} \cdot \frac{1}{(6/7)^{2}} = \frac{1}{49} \cdot \frac{49}{36} = \frac{1}{36}$.Thus,$S = \frac{1}{36} \cdot \frac{7}{6} = \frac{7}{216}$.

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

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