Question

Quantos termos da PG (1,4,16..) foram somados para obter 1365?

Answer 1

$S_n = \frac{a1 (1-q^n)}{1 - q}$


$1365 = \frac{1 (1-4^n)}{1 - 4}$


$-4095 = 1-4^n$

$4096=4^n$

$2^{12}=2^{2n}$

$12=2n$

$n=6$


Hugs