Just a variation in geometric sums

Algebra Level 3

a 0 + a 1 + a 2 + … + a n = a n + 1 − 1 a − 1 \large a^0 + a^1 + a^2 + \ldots + a^n = \frac{a^{n+1}-1}{a-1}

We know that for the geometric sum above holds true. Can you find a formula for the sum

a 0 + a 2 + a 4 + … + a 2 n ? \large a^0 + a^2 + a^4 + \ldots + a^{2n}?

(a^{2(n+1)}-1)/(a^{2}-1) (a^{2(n+1)}-1)/(a-1) (a^{2n+1)}-1)/(a^{2}-1) None of the rest (a^{2(n+1)}-1)/(a^{2n}-1)

Juan Rodriguez by Juan Rodriguez , 25, Colombia

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