Question

Calcule a soma dos termos da PG (1, 10^-1, 10^-2, 10^-3, ..., 10^-8).

Answer 1

Vamos começar determinando a razão:

$razao\,(q)~=~\frac{a_2}{a_1}\\\\\\razao\,(q)~=~\frac{10^{-1}}{1}\\\\\\\boxed{razao\,(q)~=~10^{-1}~~ou~~\frac{1}{10}}$

Agora, utilizando a equação do termo geral da PG, vamos achar o numero de termos dessa PG:

$a_n~=~a_1~.~q^{n-1}\\\\\\10^{-8}~=~1~.~\left(10^{-1}\right)^{n-1}\\\\\\\left(10^{-1}\right)^{8}~=~\left(10^{-1}\right)^{n-1}\\\\\\n-1~=~8\\\\\\\boxed{n~=~9~termos}$

Por fim, utilizando a equação da soma de termos da PG, temos:

$S_n~=~\frac{a_1~.~\left(q^n-1\right)}{q-1}\\\\\\S_9~=~\frac{1~.~\left(\left(10^{-1}\right)^9-1\right)}{10^{-1}-1}\\\\\\S_9~=~\frac{\left(\frac{1}{10^9}-1\right)}{\frac{1}{10}-1}\\\\\\S_9~=~\frac{\left(\frac{1}{10^{9}}-1\right)}{-\frac{9}{10}}\\\\\\S_9~=~\left(\frac{1}{10^{9}}-1\right)~.~-\frac{10}{9}\\\\\\S_9~=~\left(-\frac{1}{10^8}~+~10\right)~.~\frac{1}{9}\\\\\\S_9~=~\left(\frac{-1~.~1~+~10^8~.~10}{10^8}\right)~.~\frac{1}{9}$

$S_9~=~\left(\frac{-1~+~10^9}{9~.~10^8}\right)\\\\\\S_9~=~\frac{999.999.999}{900.000.000}\\\\\\\boxed{S_9~=~\frac{111.111.111}{100.000.000}~~ou~~1,11111111}$