Question

Find the sum of the geometric series with a1 = -1, r = -2 And a10 = 512

Answer 1

Assuming you want the sum of the first
`n` terms:


`S_n=a_1+a_2+\cdots+a_(n-1)+a_n`

`S_n=a_1+ra_1+\cdots+r^(n-2)a_1+r^(n-1)a_1`

`rS_n=ra_1+r^2a_1+\cdots+r^(n-1)a_1+r^na_1`


`\implies S_n-rS_n=a_1-r^na_1\iff(1-r)S_n=(1-r^n)a_1\implies S_n=(1-r^n)/(1-r)a_1`

You're given that
`a_1=-1` and
`r=-2`, so


`S_n=(1-(-2)^n)/(1-(-2))(-1)=-\frac{1-(-2)^n}3`

The fact that you're given
`a_(10)` makes me think you're supposed to find
`S_(10)`, which is just a matter of setting
`n=10` in the formula above.


`S_(10)=-\frac{1-(-2)^(10)}3=-\frac{1-1024}3=341`