find-the-value-of-n-1-n-3-3-n-

Question Number 46852 by maxmathsup by imad last updated on 01/Nov/18

f i n d t h e v a l u e o f \sum_{n = 1}^{\infty} \frac{n^{3}}{3^{n}}

Commented by maxmathsup by imad last updated on 02/Nov/18

w e h a v e p r o v e d t h a t f o r ∣ x ∣< 1 \sum_{n = 1}^{\infty} {n x}^{n} = \frac{x}{{(1 - x)}^{2}} \Rightarrow

\sum_{n = 1}^{\infty} n^{2} x^{n - 1} = \frac{{(1 - x)}^{2} + 2 x (1 - x)}{{(1 - x)}^{4}} = \frac{1 - x + 2 x}{{(1 - x)}^{3}} = \frac{x + 1}{{(1 - x)}^{3}} \Rightarrow

n^{2} x^{n} = \frac{x^{2} + x}{{(1 - x)}^{3}} \Rightarrow \sum_{n = 1}^{\infty} n^{3} x^{n - 1} = \frac{(2 x + 1) {(1 - x)}^{3} + 3 {(1 - x)}^{2} (x^{2} + x)}{{(1 - x)}^{6}}

= \frac{(2 x + 1) (1 - x) + 3 (x^{2} + x)}{{(1 - x)}^{4}} = \frac{2 x - 2 x^{2} + 1 - x + 3 x^{2} + 3 x}{{(1 - x)}^{4}} = \frac{x^{2} + 4 x + 1}{{(1 - x)}^{4}} \Rightarrow

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

x = \frac{1}{3} \Rightarrow \sum_{n = 1}^{\infty} n^{3} {(\frac{1}{3})}^{n} = \frac{\frac{1}{27} + \frac{4}{9} + \frac{1}{3}}{{(\frac{2}{3})}^{4}} = \frac{1}{27} \frac{1 + 12 + 9}{\frac{2^{4}}{3^{4}}} = \frac{1}{27} \frac{3^{4}}{2^{4}} . 22

= \frac{3}{8.2} . \frac{11.2}{1} = \frac{33}{8} \Rightarrow ★ \sum_{n = 1}^{\infty} \frac{n^{3}}{3^{n}} = \frac{33}{8} ★

Answered by Smail last updated on 02/Nov/18

\frac{1}{1 - x} = \underset{n = 0}{\sum^{\infty}} x^{n} w i t h ∣ x ∣< 1

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

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

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

\underset{n = 1}{\sum^{\infty}} n^{3} x^{n - 1} = \frac{(2 x + 1) (1 - x) + 3 x (1 + x)}{{(1 - x)}^{4}}

= \frac{x^{2} + 4 x + 1}{{(1 - x)}^{4}}

\underset{n = 1}{\sum^{\infty}} n^{3} x^{n} = \frac{x (x^{2} + 4 x + 1)}{{(1 - x)}^{4}}

x = \frac{1}{3}

\underset{n = 1}{\sum^{\infty}} \frac{n^{3}}{3^{n}} = \frac{\frac{1}{9} + \frac{4}{3} + 1}{3 {(\frac{2}{3})}^{4}} = \frac{33}{8}

Commented by maxmathsup by imad last updated on 02/Nov/18

c o r r e c t a n s w e r t h a n k s s i r .

Commented by Smail last updated on 02/Nov/18

T h a n k y o u f o r p o s t i n g q u i s t i o n s l i k e t h i s .

Facebook Tweet Pin