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**Amer** $\displaystyle P_k = \sum_{i=0}^{k} x^i = \frac{(1-x^{k+1})}{1-x} $

test $\displaystyle P_1$

$\displaystyle \sum_{i=0}^1 x^i = x^0 + x=1+x $

$\displaystyle \frac{1 - x^{1+1}}{1-x} = 1+x $

$\displaystyle P_1 $ is true

suppose that $\displaystyle P_k $ is true try

$\displaystyle P_{k+1} $ is true

$\displaystyle \sum_{i=0}^{k+1} x^i = x^{k+1}+\sum_{i=0}^{k} x^i $

but $\displaystyle \sum _{i=0}^{k} x^k = \frac{1- x^{k+1}}{1-x} $

so

$\displaystyle \sum_{i=0}^{k+1} x^i = \frac{1- x^{k+1}}{1-x} + x^{k+1} $

$\displaystyle \frac{1- x^{k+1}}{1-x} = \frac{1- x^{k+1} + x^{k+1} - x^{k+2}}{1-x} = \frac{1-x^{k+2}}{1-x} $

$\displaystyle \sum_{i=0}^{k+1} x^i = \frac{1-x^{k+2}}{1-x} $

so $\displaystyle P_{k+1}$ is true the proof end