How to evaluate the series: $ \frac {20} {(5-4)(5^2-4^2)} + \frac {20^2} {(5^2-4^2)(5^3-4^3)} + \frac {20^3} {(5^3-4^3)(5^4-4^4)} + \ldots $ How to evaluate this infinte summation?
$$ \dfrac {20} {(5-4)(5^2-4^2)} + \dfrac {20^2} {(5^2-4^2)(5^3-4^3)} + \dfrac {20^3} {(5^3-4^3)(5^4-4^4)} + \ldots $$
Telescopic cancellation is the way to go, I feel, but I don't know how to get this into telescopic form.
Hints?

We have:
$$
\frac{20}{(5−4)(5^2−4^2)}=\frac{4}{5-4}-\frac{4^2}{5^2-4^2}
$$
$$
\frac{20^2}{(5^2−4^2)(5^3−4^3)}=\frac{4^2}{5^2-4^2}-\frac{4^3}{5^3-4^3}
$$
$$
\frac{20^3}{(5^3−4^3)(5^4−4^4)}=\frac{4^3}{5^3-4^3}-\frac{4^4}{5^4-4^4}
$$
Means for gathering data through equations have
$$
\frac{20}{(5−4)(5^2−4^2)}+\frac{20^2}{(5^2−4^2)(5^3−4^3)}+\frac{20^3}{(5^3−4^3)(5^4−4^4)}+\cdots
$$
$$
=\frac{4}{5-4}-\frac{4^2}{5^2-4^2}+\frac{4^2}{5^2-4^2}-\frac{4^3}{5^3-4^3}+\frac{4^3}{5^3-4^3}-\frac{4^4}{5^4-4^4}+\cdots=\frac{4}{5-4}=4
$$
