Solve the Following Questions

Question:

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

Solution:

$a_{n+2}=2 a_{n+1}+a_{n}$, let $\sum_{n=1}^{\infty} \frac{a_{n}}{8^{n}}=P$

Divide by $8^{\mathrm{n}}$ we get

$\frac{a_{n+2}}{8^{n}}=\frac{2 a_{n+1}}{8^{n}}+\frac{a_{n}}{8^{n}}$

$\Rightarrow 64 \frac{a_{n+2}}{8^{n+2}}=\frac{16 a_{n+1}}{8^{n+1}}+\frac{a_{n}}{8^{n}}$

$64 \sum_{n=1}^{\infty} \frac{a_{n+2}}{8^{n+2}}=16 \sum_{n=1}^{\infty} \frac{a_{n+1}}{8^{n+1}}+\sum_{n=1}^{\infty} \frac{a_{n}}{8^{n}}$

$64\left(P-\frac{a_{1}}{8}-\frac{a_{2}}{8^{2}}\right)=16\left(P-\frac{a_{1}}{8}\right)+P$

$\Rightarrow 64\left(\mathrm{P}-\frac{1}{8}-\frac{1}{64}\right)=16\left(\mathrm{P}-\frac{1}{8}\right)+\mathrm{P}$

$64 \mathrm{P}-8-1=16 \mathrm{P}-2+\mathrm{P}$

$47 \mathrm{P}=7$

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