Question

Calcule a soma dos termos da P.G (2,2V5,10,10V5,50,50V5,250). por favor me expliquem como é q se faz

Answer 1

Vamos começar determinando a razão (q) da PG:

$razao~=~\frac{a_2}{a_1}\\\\\\razao~=~\frac{2\sqrt{5}}{2}\\\\\\razao~=~\frac{2\!\!\!\backslash\sqrt{5}}{2\!\!\!\backslash}\\\\\\\boxed{razao~=~\sqrt{5}}$

Podemos agora utilizar a equação da soma dos termos da PG finita. Note que a PG dada possui 7 termos, ou seja, n=7.

$S_n~=~\frac{a_1\,.\,(q^n-1)}{q-1}\\\\\\S_7~=~\frac{2\,.\,\left(\sqrt{5}^{\,7}-1\right)}{\sqrt{5}-1}\\\\\\S_7~=~\frac{2\,.\,\left(\sqrt{5^7}-1\right)}{\sqrt{5}-1}\\\\\\S_7~=~\frac{2\,.\,\left(\sqrt{5^2.5^2.5^2.5}-1\right)}{\sqrt{5}-1}$

$S_7~=~\frac{2\,.\,\left(5\,.\,5\,.\,5\,.\,\sqrt{5}\,-1\right)}{\sqrt{5}-1}\\\\\\S_7~=~\frac{2\,.\,125\sqrt{5}~-~2\,.\,1}{\sqrt{5}-1}\\\\\\S_7~=~\frac{250\sqrt{5}~-~2}{\sqrt{5}-1}\\\\\\Racionaliando~a~fracao\\\\\\S_7~=~\frac{250\sqrt{5}~-~2}{\sqrt{5}-1}~.~\frac{\sqrt{5}+1}{\sqrt{5}+1}\\\\\\S_7~=~\frac{250\sqrt{5}\,.\,\sqrt{5}~+~250\sqrt{5}\,.\,1~-~2\,.\,\sqrt{5}~-~2\,.\,1}{\sqrt{5}\,.\,\sqrt{5}~+~\sqrt{5}\,.\,1~-~1\,.\,\sqrt{5}~-~1\,.\,1}$

$S_7~=~\frac{250\,.\,5~+~250\sqrt{5}~-~2\sqrt{5}~-~2}{5~+~\sqrt{5}~-~\sqrt{5}~-~1}\\\\\\S_7~=~\frac{248\sqrt{5}~+~1248}{4}\\\\\\\boxed{S_7~=~62\sqrt{5}~+~312}$