Question

Determinati numerele naturale n pentru care n+n²+n³+...+n²⁰²⁰
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Answer 1

n+n²+n³+...+n²⁰²⁰ =

n(1 + n + n^2 + ... + n^2019) =

$n(\frac{n^{2019+1}-1 }{n-1})=\\\frac{n^{2021}-n }{n-1}$

De aici adaptezi tu la exercitiul tau, in fine formula generala este:

$\frac{a^{n+1}-1 }{a- 1}$