Question

What is the sum of the series?

4+12+36+108+...+8748


Can someone first explain the log part? When you get your log result, do you round it?

Answer 1

`t_n = a r^(n-1)`

a = 4, r = 3,
`t_n = 8748`

No rounding needed? It's a whole number


`8748 = 4 \cdot 3^(n-1) \\ \\ \displaystyle(8748)/(4) = 3^(n-1) \\ \\ 2187 = 3^(n-1) \\ \\ \log_3(2187) = n - 1 \implies n = \log_3(2187) + 1 \implies \\ \\ \\ n = (\ln 2187)/(\ln 3) + 1 = 7 + 1 = 8`


`\displaystyle S_n=(a(1-r^n))/(1-r) \implies S_8 =\displaystyle (4(1 - 3^8))/(1 - 3) \\ \\ \\ = 13120`

Sum is 13120