Question 472498
I will show you the formula plus how it is derived so that you don't blindly memorize it. Blindly memorizing formulas is a common pitfall students make, and it makes students less likely to actually "understand" the material.


Let


*[tex \LARGE S = \sum_{i=0}^{19} 3^i = 3^0+3^1+3^2+...3^{19}]


Then,


*[tex \LARGE 3S = \sum_{i=0}^{19} 3^{i+1} = 3^1 + 3^2 + ... + 3^{20}]


We can do 3S - S and obtain lots of cancellation:


*[tex \LARGE 3S - S = 2S = 3^{20} - 3^0]


*[tex \LARGE S = \frac{3^{20} - 1}{2}]


To generalize, you would let the common difference be something like "r" then the denominator of 2 would be replaced with r-1. Also, for an arbitrary exponent n (here n = 19) the 20 would be replaced with n+1.