Answered: "Calculate:∑n=0∞n(45)n+1"

Vofteldetgpt

Answered question

2022-02-28

Calculate:

\sum_{n = 0}^{\infty} n {(\frac{4}{5})}^{n + 1}

faraidz3i

Beginner2022-03-01Added 10 answers

So,

\sum_{n \geq 0} n {(\frac{4}{5})}^{n + 1} = 1 {(\frac{4}{5})}^{2} + 2 {(\frac{4}{5})}^{3} + 3 {(\frac{4}{5})}^{4} + \dots

Now,

S = 1 {(\frac{4}{5})}^{2} + 2 {(\frac{4}{5})}^{3} + 3 {(\frac{4}{5})}^{4} + \dots

(1)

\frac{5 S}{4} = 1 {(\frac{4}{5})}^{1} + 2 {(\frac{4}{5})}^{2} + 3 {(\frac{4}{5})}^{3} + \dots

Substract equation (1) from (2).

\frac{5 S}{4} - S = \frac{4}{5} + (2 - 1) {(\frac{4}{5})}^{2} + (3 - 2) {(\frac{4}{5})}^{3} + \dots

\frac{S}{4} = \frac{4}{5} + {(\frac{4}{5})}^{2} + {(\frac{4}{5})}^{3} + \dots

This is nothing but infinite GP, where

| r | < 1

r = \frac{4}{5}

So you get

\frac{S}{4} = \frac{\frac{4}{5}}{1 - \frac{4}{5}} \Rightarrow S = 16

Manraj Horton

Beginner2022-03-02Added 6 answers

Expanding the suggestions given in the comment, By the power series expansion we have that,

\frac{1}{1 - x} = \sum_{k \geq 0} x^{k}, | x | < 1

Differentiation on both sides,

\frac{1}{{(1 - x)}^{2}} = \sum_{k \geq 0} k x^{k - 1}

\frac{1}{{(1 - x)}^{2}} = \frac{1}{x} \sum_{k \geq 0} k x^{k}

Multiply by

x^{2}

on both sides,

\frac{x^{2}}{(1 - x^{2})} = \sum_{k \geq 0} k x^{k + 1}

Now according to required question,

x = \frac{4}{5}

. So answer will be

\frac{{(\frac{4}{5})}^{2}}{{(1 - \frac{4}{5})}^{2}} = \sum_{k \geq 0} k {(\frac{4}{5})}^{k + 1}

= \frac{{(\frac{4}{5})}^{2}}{{(1 - \frac{4}{5})}^{2}} = \frac{4^{2}}{1} = 16

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