find-n-1-n-n-1-3-n-

Question Number 30049 by abdo imad last updated on 15/Feb/18

f i n d \sum_{n = 1}^{\infty} \frac{n (n + 1)}{3^{n}} .

Commented by prof Abdo imad last updated on 16/Feb/18

w e h a v e p r o v e d t h a t \sum_{n = 0}^{\infty} (n + 1) x^{n} = \frac{1}{{(1 - x)}^{2}}

f o r ∣ x ∣< 1 d u e t o u n i f o r m c o n v e r g e n c e o n [0, 1]

b y d e r i v a t i o n \sum_{n = 1}^{\infty} n (n + 1) x^{n - 1} = \frac{d}{d x} ({(1 - x)}^{- 2})

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

\sum_{n = 1}^{\infty} n (n + 1) x^{n} = \frac{2 x}{{(1 - x)}^{3}} a n d f o r x = \frac{1}{3} w e g e t

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