Find-the-value-of-n-1-n-2-2-n-1-

Question Number 33088 by Joel578 last updated on 10/Apr/18

Find the value of

\underset{n = 1}{\sum^{\infty}} \frac{n^{2}}{2^{n - 1}}

Commented by abdo imad last updated on 10/Apr/18

l e t c o n s i d e r t h e s e r i e s (x) = \sum_{n = 0}^{\infty} x^{n} w i t h ∣ x ∣< 1

w e h a v e s (x) = \frac{1}{1 - x} \Rightarrow s^{^{'}} (x) = \frac{1}{{(1 - x)}^{2}} \Rightarrow \sum_{n = 1}^{\infty} n x^{n - 1}

= \frac{1}{{(1 - x)}^{2}} a l s o s^{^{″}} (x) = - \frac{2 (- 1) (1 - x)}{{(1 - x)}^{4}} = \frac{2}{{(1 - x)}^{3}} \Rightarrow

\sum_{n = 2}^{\infty} n (n - 1) x^{n - 2} = \frac{2}{{(1 - x)}^{3}} \Rightarrow

\sum_{n = 2}^{\infty} n (n - 1) x^{n - 1} = \frac{2 x}{{(1 - x)}^{3}} \Rightarrow

\sum_{n = 2}^{\infty} n^{2} x^{n - 1} = \frac{2 x}{{(1 - x)}^{3}} + \sum_{n = 2}^{\infty} {n x}^{n - 1}

= \frac{2 x}{{(1 - x)}^{3}} + \frac{1}{{(1 - x)}^{2}} - 1 \Rightarrow \sum_{n = 1}^{\infty} n^{2} x^{n - 1} = \frac{2 x}{{(1 - x)}^{3}} + \frac{1}{{(1 - x)}^{2}}

a f t e r w e t a k e x = \frac{1}{2} \Rightarrow

\sum_{n = 1}^{\infty} \frac{n^{2}}{2^{n - 1}} = \frac{1}{{(\frac{1}{2})}^{3}} + \frac{1}{{(\frac{1}{2})}^{2}} = 8 + 4 = 12 .

Commented by abdo imad last updated on 10/Apr/18

r e m a r k ★ w e h a v e 2 \sum_{n = 1}^{\infty} \frac{n^{2}}{2^{n}} = 12 \Rightarrow \sum_{n = 1}^{\infty} \frac{n^{2}}{2^{n}} = 6 .

Commented by Joel578 last updated on 13/Apr/18

T h a n k y o u v e r y m u c h

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