Answer:
u need to put pictures?
Step-by-step explanation:
.......,??????
Looks like you just evaluated the summand for the given value of
![n](https://tex.z-dn.net/?f=n)
, whereas the question is asking you to find the value of the sum for the first
![n](https://tex.z-dn.net/?f=n)
terms.
Let
![S_k=\displaystyle\sum_{n=1}^k\frac3{(-2)^n}](https://tex.z-dn.net/?f=S_k%3D%5Cdisplaystyle%5Csum_%7Bn%3D1%7D%5Ek%5Cfrac3%7B%28-2%29%5En%7D)
. Then
![S_k](https://tex.z-dn.net/?f=S_k)
is the
![k](https://tex.z-dn.net/?f=k)
th partial sum.
![S_1](https://tex.z-dn.net/?f=S_1)
happens to be the first term in the series, which is why that box is marked correct:
![S_1=\displaystyle\sum_{n=1}^1\frac3{(-2)^n}=\frac3{(-2)^1}=-1.5](https://tex.z-dn.net/?f=S_1%3D%5Cdisplaystyle%5Csum_%7Bn%3D1%7D%5E1%5Cfrac3%7B%28-2%29%5En%7D%3D%5Cfrac3%7B%28-2%29%5E1%7D%3D-1.5)
But the next partial sum is not correct:
![S_2=\displaystyle\sum_{n=1}^2\frac3{(-2)^n}=\frac3{(-2)^1}+\frac3{(-2)^2}=-0.75](https://tex.z-dn.net/?f=S_2%3D%5Cdisplaystyle%5Csum_%7Bn%3D1%7D%5E2%5Cfrac3%7B%28-2%29%5En%7D%3D%5Cfrac3%7B%28-2%29%5E1%7D%2B%5Cfrac3%7B%28-2%29%5E2%7D%3D-0.75)
and this is not the same notion as the second term (which indeed is 0.75) in the series.
Ok what is the question lol
![\frac{12}{100}=0.12](https://tex.z-dn.net/?f=%5Cfrac%7B12%7D%7B100%7D%3D0.12)
To get the 0.12 to repeat, the trick is to subtract 1 from the denominator
![\frac{12}{99}=0.12121212...](https://tex.z-dn.net/?f=%5Cfrac%7B12%7D%7B99%7D%3D0.12121212...)
Then reduce the fraction.
![\frac{12}{99}={4}{33}](https://tex.z-dn.net/?f=%5Cfrac%7B12%7D%7B99%7D%3D%7B4%7D%7B33%7D)
now add the 3 to get