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Prove by mathematical induction: 1+2+2^2+2^3+....+2^n-1=2^n-1

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First show the statement holds for the base case (presumably
n=1):


\displaystyle\sum_(i=1)^1 2^(i-1)=2^(1-1)=2^0=1

Meanwhile, the right hand side evaluates to
2^1-1=2-1=1, so the base case holds.

Now assume the formula holds for
n=k; that is,


\displaystyle\sum_(i=1)^k 2^(i-1)=2^0+2^1+\cdots+2^(k-1)=2^k-1

and use this hypothesis to show that the formula holds for
n=k+1. You have


\displaystyle\sum_(i=1)^(k+1)2^(i-1)=2^k+\sum_(i=1)^k 2^(i-1)=2^k+2^k-1=2*2^k-1=2^(k+1)-1

so the formula holds and the proof is complete.
User Daniel Steinmann
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