Let {a_{n}}_{n=1}^{infty} be a sequence such | vedclass

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(B) Let $P = \sum_{n=1}^{\infty} \frac{a_{n}}{8^{n}}$.Given the recurrence relation $a_{n+2} = 2a_{n+1} + a_{n}$.Dividing by $8^{n+2}$,we get $\frac{a

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$7$

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(B) Let $P = \sum_{n=1}^{\infty} \frac{a_{n}}{8^{n}}$.Given the recurrence relation $a_{n+2} = 2a_{n+1} + a_{n}$.Dividing by $8^{n+2}$,we get $\frac{a_{n+2}}{8^{n+2}} = \frac{2a_{n+1}}{8^{n+2}} + \frac{a_{n}}{8^{n+2}}$.Summing from $n=1$ to $\infty$:$\sum_{n=1}^{\infty} \frac{a_{n+2}}{8^{n+2}} = \frac{2}{8} \sum_{n=1}^{\infty} \frac{a_{n+1}}{8^{n+1}} + \frac{1}{64} \sum_{n=1}^{\infty} \frac{a_{n}}{8^{n}}$.Let $S = \sum_{n=1}^{\infty} \frac{a_{n}}{8^{n}} = P$. Then $\sum_{n=1}^{\infty} \frac{a_{n+1}}{8^{n+1}} = P - \frac{a_{1}}{8} = P - \frac{1}{8}$.And $\sum_{n=1}^{\infty} \frac{a_{n+2}}{8^{n+2}} = P - \frac{a_{1}}{8} - \frac{a_{2}}{64} = P - \frac{1}{8} - \frac{1}{64}$.Substituting these into the equation:$P - \frac{1}{8} - \frac{1}{64} = \frac{1}{4}(P - \frac{1}{8}) + \frac{1}{64}P$.Multiplying by $64$:$64P - 8 - 1 = 16(P - \frac{1}{8}) + P$.$64P - 9 = 16P - 2 + P$.$64P - 9 = 17P - 2$.$47P = 7$.

Let $\{a_{n}\}_{n=1}^{\infty}$ be a sequence such that $a_{1}=1, a_{2}=1$ and $a_{n+2}=2a_{n+1}+a_{n}$ for all $n \geq 1$. Then the value of $47 \sum_{n=1}^{\infty} \frac{a_{n}}{2^{3n}}$ is equal to $.....$

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