Prove the following by using the principle of mathematical induction for all $n \in N:$
$1+2+2^{2}+\ldots+2^{n}=2^{n+1}-1$
$1+2+2^{2}+\ldots+2^{n}=2^{n+1}-1$
(N/A) Let $P(n): 1+2+2^2+\ldots+2^n = 2^{n+1}-1$ for all $n \in N$.
Step $I$: For $n=1$,
$LHS$ $= 1+2^1 = 3$.
$RHS$ $= 2^{1+1}-1 = 2^2-1 = 4-1 = 3$.
Since $LHS$ $= RHS$,the statement is true for $n=1$.
Step $II$: Assume $P(k)$ is true for some $k \in N$,i.e.,$1+2+2^2+\ldots+2^k = 2^{k+1}-1$.
Step $III$: For $n=k+1$,we need to show $P(k+1)$ is true,i.e.,$1+2+2^2+\ldots+2^k+2^{k+1} = 2^{(k+1)+1}-1 = 2^{k+2}-1$.
$LHS$ $= (1+2+2^2+\ldots+2^k) + 2^{k+1}$.
Using the assumption from Step $II$,$LHS$ $= (2^{k+1}-1) + 2^{k+1}$.
$= 2 \times 2^{k+1} - 1 = 2^{k+2}-1$.
Since $LHS$ $= RHS$,the statement is true for $n=k+1$.
By the principle of mathematical induction,the statement is true for all $n \in N$.
Step $I$: For $n=1$,
$LHS$ $= 1+2^1 = 3$.
$RHS$ $= 2^{1+1}-1 = 2^2-1 = 4-1 = 3$.
Since $LHS$ $= RHS$,the statement is true for $n=1$.
Step $II$: Assume $P(k)$ is true for some $k \in N$,i.e.,$1+2+2^2+\ldots+2^k = 2^{k+1}-1$.
Step $III$: For $n=k+1$,we need to show $P(k+1)$ is true,i.e.,$1+2+2^2+\ldots+2^k+2^{k+1} = 2^{(k+1)+1}-1 = 2^{k+2}-1$.
$LHS$ $= (1+2+2^2+\ldots+2^k) + 2^{k+1}$.
Using the assumption from Step $II$,$LHS$ $= (2^{k+1}-1) + 2^{k+1}$.
$= 2 \times 2^{k+1} - 1 = 2^{k+2}-1$.
Since $LHS$ $= RHS$,the statement is true for $n=k+1$.
By the principle of mathematical induction,the statement is true for all $n \in N$.
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