Another Summation

Algebra Level 5

$\large\displaystyle\sum_{n=8}^{ \infty}\dfrac{1}{(n-4)(n-2)n(n+2)(n+4)}$

The summation above is equal to $\dfrac{a}{b}$ , where $\text{gcd}(a,b) = 1$ . Find $a+b$ .

1 solution

Ankit Kumar Jain
May 26, 2017

$\displaystyle \sum_{n=8}^{\infty} \dfrac{1}{(n-4)(n-2)n(n+2)(n+4)}$

$\Rightarrow \displaystyle \sum_{n=8}^{\infty} \dfrac18\cdot \dfrac{(n+4) - (n-4)}{(n-4)(n-2)n(n+2)(n+4)}$

$\Rightarrow \displaystyle \sum_{n=8}^{\infty} \dfrac18\cdot\left(\dfrac1{(n-4)(n-2)n(n+2)} - \dfrac1{(n-2)n(n+2)(n+4)}\right)$

$\Rightarrow \dfrac18\cdot\left(\dfrac1{4\cdot6\cdot8\cdot10} + \dfrac1{5\cdot7\cdot9\cdot11}\right)$

$\Rightarrow \boxed{\dfrac{359}{3548160}}$

did the same way

I Gede Arya Raditya Parameswara - 3 years, 5 months ago