Question

Calcule a soma dos onze primeiros termos da pg (3;9/2;27/4;...).

Answer 1

Resposta:

Explicação passo-a-passo:

Temos que a razão q é:

q = a₂ ÷ a₁

q = 9/2 ÷ 3

q = 9/2 × 1/3

q = 3/2

$S_{11}=\frac{a_{1}[(q^{11-1})-1]}{q-1}=>S_{11}=\frac{3.[(\frac{3}{2})^{10}-1]}{\frac{3}{2}-1}=>S_{11}=\frac{\frac{59049}{1024}-1}{\frac{1}{2}}=>S_{11}=\frac{\frac{59049-1024}{1024}}{\frac{1}{2}}=>S_{11}=\frac{\frac{58025}{1024}}{\frac{1}{2}}=>S_{11}=\frac{58025}{1024}.\frac{2}{1}=>S_{11}=\frac{58025}{512}$