Calculate-n-1-n-2-3-n-

Question Number 141791 by Chhing last updated on 23/May/21

Calculate

\underset{n = 1}{\sum^{+ \infty}} \frac{n^{2}}{3^{n}}

Answered by mr W last updated on 23/May/21

\underset{n = 1}{\sum^{\infty}} x^{n} = x + x^{2} + x^{3} + \dots = \frac{x}{1 - x}

\underset{n = 1}{\sum^{\infty}} {n x}^{n - 1} = \frac{1}{1 - x} + \frac{x}{{(1 - x)}^{2}}

\underset{n = 1}{\sum^{\infty}} {n x}^{n} = \frac{x}{1 - x} + \frac{x^{2}}{{(1 - x)}^{2}}

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

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

w i t h x = \frac{1}{3} :

\Rightarrow \underset{n = 1}{\sum^{\infty}} \frac{n^{2}}{3^{n}} = \frac{1}{2} + \frac{3}{4} + \frac{2}{8} = \frac{3}{2}

Answered by qaz last updated on 23/May/21

\underset{n = 1}{\sum^{\infty}} \frac{n^{2}}{3^{n}} = \underset{n = 0}{\sum^{\infty}} \frac{{(n + 1)}^{2}}{3^{n + 1}} = \frac{1}{3} {(x D + 1)}^{2} ∣_{x = 1 / 3} \frac{1}{1 - x}

= \frac{1}{3} {[{(x D)}^{2} + 2 x D + 1]}_{x = 1 / 3} \frac{1}{1 - x}

= \frac{1}{3} (x^{2} D^{2} + 3 x D + 1) ∣_{x = 1 / 3} \frac{1}{1 - x}

= \frac{1}{3} {[\frac{2 x^{2}}{{(1 - x)}^{3}} + \frac{3 x}{{(1 - x)}^{2}} + \frac{1}{1 - x}]}_{x = 1 / 3}

= \frac{1}{3} (\frac{3}{4} + \frac{9}{4} + \frac{3}{2})

= \frac{3}{2}

Answered by mathmax by abdo last updated on 23/May/21

let s (x) = \sum_{n = 1}^{\infty} x^{n} with [∣ x ∣< 1 \Rightarrow s^{^{'}} (x) = \sum_{n = 1}^{\infty} {nx}^{n - 1} \Rightarrow

{xs}^{^{'}} (x) = \sum_{n = 1}^{\infty} {nx}^{n} \Rightarrow s^{^{'}} (x) + {xs}^{(2)} (x) = \sum_{n = 1}^{\infty} n^{2} x^{n - 1} \Rightarrow

{xs}^{^{'}} (x) + x^{2} s^{(2)} (x) = \sum_{n = 1}^{\infty} n^{2} x^{n}

x = \frac{1}{3} \Rightarrow \sum_{n = 1}^{\infty} \frac{n^{2}}{3^{n}} = \frac{1}{3} s^{^{'}} (\frac{1}{3}) + \frac{1}{9} s^{^{″}} (\frac{1}{3}) we have

s (x) = \frac{1}{1 - x} \Rightarrow s^{^{'}} (x) = \frac{1}{{(1 - x)}^{2}} and s^{″} (x) = - \frac{2 (1 - x) (- 1)}{{(1 - x)}^{4}}

= \frac{2}{{(1 - x)}^{3}} so s^{^{'}} (\frac{1}{3}) = \frac{1}{{(\frac{2}{3})}^{2}} = \frac{9}{4} and s^{(2)} (\frac{1}{3}) = 2 \times \frac{27}{8} = \frac{27}{4} \Rightarrow

\sum_{n = 1}^{\infty} \frac{n^{2}}{3^{n}} = \frac{1}{3} . \frac{9}{4} + \frac{1}{9} . \frac{27}{4} = \frac{3}{4} + \frac{3}{4} = \frac{6}{4} = \frac{3}{2}

Answered by qaz last updated on 23/May/21

\underset{n = 0}{\sum^{\infty}} \frac{{(n + 1)}^{2}}{3^{n + 1}} = \underset{n = 0}{\sum^{\infty}} \frac{{(n + 1)}^{2}}{Γ (n + 1)} \int_{0}^{\infty} u^{n} e^{- 3 u} d u

= \int_{0}^{\infty} e^{- 3 u} \underset{n = 0}{\sum^{\infty}} \frac{{(n + 1)}^{2}}{n!} u^{n} d u

= \int_{0}^{\infty} e^{- 3 u} {(x D + 1)}^{2} ∣_{x = u} \underset{n = 0}{\sum^{\infty}} \frac{x^{n}}{n!} d u

= \int_{0}^{\infty} e^{- 3 u} (x^{2} D^{2} + 3 x D + 1) ∣_{x = u} e^{x} d u

= \int_{0}^{\infty} e^{- 3 u} (x^{2} + 3 x + 1) e^{x} ∣_{x = u} d u

= \int_{0}^{\infty} e^{- 2 u} (u^{2} + 3 u + 1) d u

= \frac{2}{2^{3}} + \frac{3}{2^{2}} + \frac{1}{2}

= \frac{3}{2}

i h a v e f o u n d a n o t h e r i n t r e s t i n g w a y .

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