2022/027) If $\sum_{k=1}^n \frac{2k+1}{(k^2+k)^2} = .9999$ find n

 We have

$\frac{2k+1}{(k^2+k)^2} = \frac{(k+1)^2 - k^2}{(k+1)^2 k^2} =  \frac{1}{k^2} - \frac{1}{(k+1)^2}$

The above is a telescopic sum and we have

 $\sum_{k=1}^n \frac{2k+1}{(k^2+k)^2} = \frac{1}{1} - \frac{1}{(n+1)^2} = .9999$

so  $\frac{1}{(n+1)^2} = 1- .9999 = .0001 = \frac{1}{100^2}$

n+ 1 = 100 or n = 99