A brilliant sum

Calculus Level 5

$\alpha= \displaystyle \sum_{b=1}^{\infty}\left(\displaystyle \sum_{r=1}^{\infty}\left(\displaystyle \sum_{i=1}^{\infty}\left(\displaystyle \sum_{l=1}^{\infty}\left(\displaystyle \sum_{a=1}^{\infty}\left(\displaystyle \sum_{n=1}^{\infty}\left(\displaystyle \sum_{t=1}^{\infty}\left(\dfrac{1}{x^{b+r+i+l+l+i+a+n+t}}\right)\right)\right)\right)\right)\right)\right)$

If $\alpha=\dfrac{1}{(x+a)^b(x+c)^d}$ find $|a|+|b|+|c|+|d|$

The answer is 11.

1 solution

Rishabh Jain
Jun 11, 2016

$\large\sum_{j=1}^\infty \frac{1}{x^j}=\frac{1}{x-1} \\ (\text{where}~j=b,r,a,n,t)$ $\begin{aligned}\sum_{i=1}^{\infty} \frac{1}{x^{2i}} = \sum_{l=1}^{\infty} \frac{1}{x^{2l}} =&\frac{1}{x^2-1}\\=& \dfrac1{(x-1)(x+1)}\end{aligned}$ Hence,

$\begin{aligned} \alpha=&\left(\dfrac1{x-1}\right)^5\left(\dfrac1{(x-1)(x+1)}\right)^2\\ =&\dfrac{1}{(x-1)^7(x+1)^2}\end{aligned}$

$\large 1+7+1+2=\boxed{\color{#D61F06}{11}}$

Very nice solution and very simple.

You could simplify it just a bit by removing some of the summations from the top and replacing the b with some abstract letter like j.

$\sum_{b=1}^\infty \frac{1}{x^b}=\sum_{r=1}^\infty \frac{1}{x^r}=\sum_{a=1}^\infty \frac{1}{x^a}=\sum_{n=1}^\infty \frac{1}{x^n}=\sum_{t=1}^\infty \frac{1}{x^t}=\frac{1}{x-1}$

Can just be

$\sum_{j=1}^\infty \frac{1}{x^j}=\frac{1}{x-1}$

Trevor Arashiro - 5 years ago

Done.... Thanks.. :-)

Rishabh Jain - 5 years ago

A brilliant sum

The answer is 11.

1 solution

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