Sum 'em Up

Calculus Level 3

5 1 4 2 + 11 4 2 7 2 + 17 7 2 1 0 2 + \frac 5{1\cdot 4^2}+\frac{11}{4^2\cdot7^2}+\frac{17}{7^2\cdot10^2}+\cdots

Find the value of the closed form of the above series to 2 decimal places.


The answer is 0.33.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

Chew-Seong Cheong
Jan 29, 2017

S = 5 1 4 2 + 11 4 2 7 2 + 17 7 2 1 0 2 + = n = 0 ( 3 n + 1 ) + ( 3 n + 4 ) ( 3 n + 1 ) 2 ( 3 n + 4 ) 2 = n = 0 ( 3 n + 4 ) ( 3 n + 1 ) ( 3 n + 4 ) ( 3 n + 1 ) ( 3 n + 4 ) + ( 3 n + 1 ) ( 3 n + 1 ) 2 ( 3 n + 4 ) 2 = n = 0 1 3 ( 3 n + 4 ) 2 ( 3 n + 1 ) 2 ( 3 n + 1 ) 2 ( 3 n + 4 ) 2 = 1 3 n = 0 ( 1 ( 3 n + 1 ) 2 1 ( 3 n + 4 ) 2 ) = 1 3 ( n = 0 1 ( 3 n + 1 ) 2 n = 1 1 ( 3 n + 1 ) 2 ) = 1 3 1 = 1 3 0.33 \begin{aligned} S & = \frac 5{1\cdot 4^2} + \frac {11}{4^2\cdot 7^2} + \frac {17}{7^2\cdot 10^2} + \cdots \\ & = \sum_{n=0}^\infty \frac {(3n+1)+(3n +4)}{(3n+1)^2(3n +4)^2} \\ & = \sum_{n=0}^\infty \frac {(3n+4)-(3n+1)}{(3n+4)-(3n+1)} \cdot \frac {(3n+4)+(3n +1)}{(3n+1)^2(3n +4)^2} \\ & = \sum_{n=0}^\infty \frac 13 \cdot \frac {(3n+4)^2-(3n +1)^2}{(3n+1)^2(3n +4)^2} \\ & = \frac 13 \sum_{n=0}^\infty \left(\frac 1{(3n+1)^2}-\frac 1{(3n+4)^2}\right) \\ & = \frac 13 \left(\sum_{n=0}^\infty \frac 1{(3n+1)^2} - \sum_{\color{#D61F06}n=1}^\infty \frac 1{\color{#D61F06}(3n+1)^2} \right) \\ & = \frac 13 \cdot 1 = \frac 13 \approx \boxed{0.33} \end{aligned}

Sravanth C.
Jan 29, 2017

We can write the above summation as k = 0 ( 3 k + 1 ) + ( 3 k + 4 ) ( 3 k + 1 ) 2 ( 3 k + 4 ) 2 = k = 0 1 ( 3 k + 1 ) 2 ( 3 k + 1 ) + 1 ( 3 k + 1 ) ( 3 k + 4 ) 2 = 1 3 k = 0 ( 3 k + 4 ) ( 3 k + 1 ) ( 3 k + 1 ) 2 ( 3 k + 4 ) + ( 3 k + 4 ) ( 3 k + 1 ) ( 3 k + 1 ) ( 3 k + 4 ) 2 = 1 3 k = 0 1 ( 3 k + 1 ) 2 1 ( 3 k + 4 ) 2 = 1 3 ( 1 1 ( 3 ( ) + 4 ) 2 ) = 1 3 \begin{aligned} \displaystyle\sum_{k=0}^{\infty}\frac{(3k+1)+(3k+4)}{(3k+1)^2(3k+4)^2}&=\displaystyle\sum_{k=0}^{\infty}\frac{1}{(3k+1)^2(3k+1)}+\frac 1{(3k+1)(3k+4)^2}\\ &=\frac 13\displaystyle\sum_{k=0}^{\infty}\frac{(3k+4)-(3k+1)}{(3k+1)^2(3k+4)}+\frac{(3k+4)-(3k+1)}{(3k+1)(3k+4)^2}\\ &=\frac 13\displaystyle\sum_{k=0}^{\infty}\frac{1}{(3k+1)^2}-\frac 1{(3k+4)^2}\\ &=\frac 13\left(1-\frac 1{(3(\infty)+4)^2}\right)\\ &=\frac 13\\ \end{aligned}

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...