A calculus problem by Sahil Silare

Calculus Level 4

s = n = 1 ( n 2 + 2 n + 3 2 n ) \large s = \sum^\infty_{n=1}\left(\frac{n^2+2n+3}{2^n}\right)

Compute the value of 1 3 + 2 3 + 3 3 + + s 3 1^3 + 2^3 + 3^3 + \cdots + s^3 .


The answer is 8281.

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2 solutions

Using, n = 1 n 2 x n = x ( x + 1 ) ( 1 x ) 3 , x < 1 \displaystyle \sum_{n=1}^{\infty} n^2 x^n = \frac{x(x+1)}{(1-x)^3} , |x|<1 & n = 1 n x n = x ( 1 x ) 2 , x < 1 \displaystyle \sum_{n=1}^{\infty} nx^n = \frac{x}{(1-x)^2} , |x|<1 we have ,

S = n = 1 n 2 2 n + 2 n 2 n + 3 1 2 n = 6 + 4 + 3 = 13 \displaystyle S=\sum_{n=1}^{\infty} \frac{n^2}{2^n} + 2\frac{n}{2^n}+3\frac{1}{2^n} = 6+4+3=13

Thereferore , r = 1 s r 3 = 8281 \displaystyle \sum_{r=1}^{s} r^3 = \boxed{8281}

How did you write the first 2 equations?

Devang Patil - 3 years, 3 months ago

n = 1 n 2 + 2 n + 3 2 n = n = 2 n 2 + 2 2 n 1 = n = 1 n 2 2 n 1 n = 1 1 n 2 2 n 1 + n = 1 2 2 n 1 n = 1 1 2 2 n 1 = 12 1 + 4 2 = 13. = n = 1 13 n 3 = { 13 ( 13 + 1 ) 2 } 2 = 8281. \begin{aligned}\\ \displaystyle~&\sum_{n=1}^{\infty}~\dfrac{n^2+2n+3}{ 2^n}\\ \displaystyle~=&\sum_{n=2}^{\infty}~\dfrac{n^2+2} {2^{n-1}}\\ \displaystyle~=&\sum_{n=1}^{\infty}~\dfrac{n^2}{ 2^{n-1}}- \sum_{n=1}^1~\dfrac{n^2}{ 2^{n-1}}+\sum_{n=1}^{\infty}~\dfrac 2 { 2^{n-1}} - \sum_{n=1}^1~\dfrac 2 { 2^{n-1}}\\ =&12-1+4-2=13.\\ \displaystyle~=&\sum_{n=1}^{13}n^3\\ =& \left \{ \dfrac{13(13+1)} 2 \right \}^2\\ =& \Large~~\color{#D61F06}{8281}.\\ \end{aligned}

n = 1 n 2 x n = x ( x + 1 ) ( 1 x ) 3 W h e n x = 2 1 , n = 1 2 n 2 x n 1 = 2 x ( x + 1 ) ( 1 x ) 3 = 2 1 2 ( 1 2 + 1 ) ( 1 1 2 ) 3 = 2 1 2 3 2 1 8 = 12 \displaystyle\\ \sum_{n=1}^{\infty} n^2*x^n=\dfrac{x(x+1)}{(1-x)^3}\\ When~x=2^{-1},\\ \sum_{n=1}^{\infty} 2*n^2*x^{n-1}=2*\dfrac{x(x+1)}{(1-x)^3} \\ = 2*\dfrac {\frac1 2*(\frac1 2+1)}{(1-\frac 1 2)^3}\\ =2*\dfrac{\frac1 2*\frac3 2}{\frac 1 8}=12
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