find the sum of $\displaystyle \Sigma_{n=1}^{\infty} \frac{2}{3} ( {\frac{-1}{4}}^{n+1} ) $
help please am I allowed to multiply by $\displaystyle \frac{-1}{4}^{-2}$
find the sum of $\displaystyle \Sigma_{n=1}^{\infty} \frac{2}{3} ( {\frac{-1}{4}}^{n+1} ) $
help please am I allowed to multiply by $\displaystyle \frac{-1}{4}^{-2}$
Hello, Intsecxtanx!
I hope I've guessed the problem correctly . . .
Find the sum of: .$\displaystyle \sum _{n=1}^{\infty} \frac{2}{3}\left(\text{-}\frac{1}{4}\right)^{n+1} $
am I allowed to multiply by $\displaystyle \left(\text{-}\frac{1}{4}\right)^{-2}$
Well, sort of . . . but why would you do that?
We have: .$\displaystyle S \;=\;\frac{2}{3}\!\left(\text{-}\frac{1}{4}\right)^2 + \frac{2}{3}\!\left(\text{-}\frac{1}{4}\right)^3 + \frac{2}{3}\!\left(\text{-}\frac{1}{4}\right)^4 + \frac{2}{3}\!\left(\text{-}\frac{1}{4}\right)^5 + \hdots $
We have a geometric series with first term, $\displaystyle a \,=\,\frac{2}{3}\left(\text{-}\frac{1}{4}\right)^2\,=\,\frac{1}{24}$
. . and common ratio, $\displaystyle r \,=\,\text{-}\frac{1}{4}$
Substitute into the formula: .$\displaystyle S \;=\;\frac{a}{1-r}$