Question

Les comparto un ejercicio muy interesante

Álgebra..............​

Answer 1

$\displaystyle\\S=\dfrac{1}{3}\sum_{n=1}^{100}\dfrac{1}{(2n+5)(2n+3)}\\ \\ \\S=-\dfrac{1}{6}\sum_{n=1}^{100}\dfrac{1}{2n+5}-\dfrac{1}{2n+3}\\ \\ \\S=-\dfrac{1}{6}\sum_{n=1}^{100}f(n)-f(n-1)=-\dfrac{1}{6}[f(100)-f(0)]\\ \\ \\\text{Donde }f(n)=\dfrac{1}{2n+5} \\ \\ \\S=-\dfrac{1}{6}\left(\dfrac{1}{205}-\dfrac{1}{5}\right)\\ \\ \\\boxed{S=\dfrac{4}{123}}$