Let ann-1- be a sequence such that a1-1- a2-1 and an-2-2an-1-an for all n- 1- Then the value of 47-n-1-an23n is equal to

A_n+2=2a_n+1+a_n nbsp;has its characteristic equation as nbsp; x^2=2x+1⇒ x=1±√(2) So, a_n =a ( 1+√(2) )^n 1+b ( 1 √(2) )^n 1 ∵ nbsp;a_1=1⇒ a+b=1 and a_2=1 ...

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Standard XII

Mathematics

Sigma n

Let ann=1∞ be...

Question

Let ${a_{n}}_{n}^{\infty}$ 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 $47 \infty \sum n = 1 \frac{a_{n}}{2^{3 n}}$ is equal to

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Solution

$a_{n + 2} = 2 a_{n + 1} + a_{n}$ has its characteristic equation as
$x^{2} = 2 x + 1 \Rightarrow x = 1 \pm \sqrt{2}$
So, $a_{n} = a {(1 + \sqrt{2})}^{n - 1} + b {(1 - \sqrt{2})}^{n - 1}$
$∵ a_{1} = 1 \Rightarrow a + b = 1$
and $a_{2} = 1 \Rightarrow (a + b) + \sqrt{2} (a - b) = 1$
$\Rightarrow a = \frac{1}{2} and b = \frac{1}{2}$
So, $a_{n} = \frac{{(1 + \sqrt{2})}^{n - 1} + {(1 - \sqrt{2})}^{n - 1}}{2}$

$\infty \sum n = 1 \frac{a_{n}}{2^{3 n}} = \frac{1}{16} ⎡ ⎣ \infty \sum n = 1 {(\frac{1 + \sqrt{2}}{8})}^{n - 1} + \infty \sum n = 1 {(\frac{1 - \sqrt{2}}{8})}^{n - 1} ⎤ ⎦$
$= \frac{1}{16} [\frac{8}{7 - \sqrt{2}} + \frac{8}{7 + \sqrt{2}}]$
$= \frac{7}{47}$

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Let {an}∞n=1 be a sequence such that a1=1,a2=1 and an+2=2an+1+an for all n≥1. Then the value of 47∞∑n=1an23n is equal to

Let ${a_{n}}_{n}^{\infty}$ 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 $47 \infty \sum n = 1 \frac{a_{n}}{2^{3 n}}$ is equal to