obter a soma dos 10 primeiros termos da P.G(6,24,96,...)
Soluções para a tarefa
Respondido por
1
Os dados são..
![\begin{cases}q=(a_2)/(a_1)=24/6=4\\
a_1=6\\
n=10~termos\\
S_{10}=?\end{cases} \begin{cases}q=(a_2)/(a_1)=24/6=4\\
a_1=6\\
n=10~termos\\
S_{10}=?\end{cases}](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7Dq%3D%28a_2%29%2F%28a_1%29%3D24%2F6%3D4%5C%5C%0Aa_1%3D6%5C%5C%0An%3D10%7Etermos%5C%5C%0AS_%7B10%7D%3D%3F%5Cend%7Bcases%7D)
Pela fórmula da P.G. finita..
![S_n= \dfrac{a_1\cdot(q^n-1)}{q-1}\\\\\\\Rightarrow S_{10}= \dfrac{6\cdot(4^{10}-1)}{4-1}= \dfrac{6\cdot(1.048.576-1)}{3}= \dfrac{6\cdot1.048.575}{3}=2.097.150 S_n= \dfrac{a_1\cdot(q^n-1)}{q-1}\\\\\\\Rightarrow S_{10}= \dfrac{6\cdot(4^{10}-1)}{4-1}= \dfrac{6\cdot(1.048.576-1)}{3}= \dfrac{6\cdot1.048.575}{3}=2.097.150](https://tex.z-dn.net/?f=S_n%3D+%5Cdfrac%7Ba_1%5Ccdot%28q%5En-1%29%7D%7Bq-1%7D%5C%5C%5C%5C%5C%5C%5CRightarrow+S_%7B10%7D%3D+%5Cdfrac%7B6%5Ccdot%284%5E%7B10%7D-1%29%7D%7B4-1%7D%3D+%5Cdfrac%7B6%5Ccdot%281.048.576-1%29%7D%7B3%7D%3D+%5Cdfrac%7B6%5Ccdot1.048.575%7D%7B3%7D%3D2.097.150)
Pela fórmula da P.G. finita..
Perguntas interessantes