Prove that $\sum_{i=0}^{n} x ^{i} = \frac{ 1-x ^{n+1} }{ 1-x }$ to be used by directly manipulating the sum: let A be the sum, and show that xA = A + x^(n+1) -1
I don't get how its going to equal $\frac{ 1-x ^{n+1} }{ 1-x }$ $xA=x\sum_{i=0}^n x^i=x(x^0+x^1+x^2+x^3+...)=x^1+x^2+x^3+x^4+...$ so then i have $\sum_{i=0}^{n} x ^{n+1}-1$
I'm stock on how its going to equal one to each other.