<s>[INST] 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?
 [/INST] Notice that:
$$\frac{20^k}{(5^k-4^k)(5^{k+1}-4^{k+1})}=\frac{4^k}{5^k-4^k}-\frac{4^{k+1}}{5^{k+1}-4^{k+1}},$$
in order to prove that:
$$\sum_{k=1}^{+\infty}\frac{20^k}{(5^k-4^k)(5^{k+1}-4^{k+1})} = 4.$$
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