Question

What is the sum of the series?

Answer 1

$t_n = a r^{n-1}$

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

No rounding needed? It's a whole number

$8748 = 4 \cdot 3^{n-1} \\ \\ \displaystyle\frac{8748}{4} = 3^{n-1} \\ \\ 2187 = 3^{n-1} \\ \\ \log_3(2187) = n - 1 \implies n = \log_3(2187) + 1 \implies \\ \\ \\ n = \frac{\ln 2187}{\ln 3} + 1 = 7 + 1 = 8$

$\displaystyle S_n=\frac{a(1-r^n)}{1-r} \implies S_8 =\displaystyle \frac{4(1 - 3^8)}{1 - 3} \\ \\ \\ = 13120$

Sum is 13120