Question

Find the sum of the series:
3^0 + 3^1 + 3^2 + 3^3 + 3^4 + ..... + 3^100

Answer 1

\[\sum_{n = 0}^{k}r^n=\frac{1-r^{k+1}}{1-r}=\frac{1-3^{101}}{1-(100)}\]

Answer 2

that isn't an answer choice, it says the answer is
(3^101 - 1)/2

Answer 3

Okay so:
\[\frac{1-r^{k+1}}{1-r}=\frac{r^{k+1}-1}{r-1}=\frac{3^{101}-1}{3-1}=\frac{3^{101}-1}{2}\]
I meant 1-3 not 1-100

Answer 4

I don't understand any of that, but thanks for the general formula, so if it were 4 instead of 3 as the base on all of those it would be?
(4^101 - 1)/3

Answer 5

Yes, all I did was pull out the negative sign from the top and distribute it into the bottom.