"Proof:∑n=0∞n2+3n+24n=12827Given hint: (n2+3n+2)=(n+2)(n+1)" was answered by our community

cooopt392a0i

Answered question

2022-02-22

Proof:

\sum_{n = 0}^{\infty} \frac{n^{2} + 3 n + 2}{4^{n}} = \frac{128}{27}

Given hint:

(n^{2} + 3 n + 2) = (n + 2) (n + 1)

legertopdxa

Beginner2022-02-23Added 8 answers

Using the formula for the sum of a geometric series:

\sum_{k = 0}^{\infty} x^{k} = \frac{1}{1 - x}

(1)
Taking the derivative of (1) and tossing the term which are 0:

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

(2)
Taking the derivative of (2) and tossing the terms which are 0:

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

(3)
Reindexing the sum in (3):

\sum_{k = 0}^{\infty} (k + 2) (k + 1) x^{k} = \frac{2}{{(1 - x)}^{3}}

Plug in

x = \frac{1}{4}

\sum_{k = 0}^{\infty} \frac{(k + 2) (k + 1)}{4^{k}} = \frac{2}{{(\frac{3}{4})}^{3}}

= \frac{128}{27}

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