Question

What is the sum of the series? 4+12+36+108+...+8748 A. 8908 B. 10,204 C. 13,120 D. 17,816

Answer 1

C

Explanation:

We know the first term is `4`, so `a=4`. Each term is 3 times bigger than the last, meaning we have `ar^(n-1)`, with `r=3`

So, we know the series follows `4(3)^(n-1)`

Foe a geometric series:
`S_n=a_1((1-r^n)/(1-r))`

We need `n` for the last term:
`4(3)^(n-1)=8748`

`3^(n-1)=2187`

`n-1=log_3(2187)=ln(2187)/ln(3)=7`

`n=7+1=8`

`S_8=4((1-3^8)/(1-3))=13120-=C`