Question

Prove by the Principle of Mathematical Induction:$1+2+2^2+...+2^n=2^{n+1}-1$ for all natural numbers n.

Answer 1

Assume given statement

Let given statement be

$P\left(n\right),i.e.,P\left(n\right)=1+2+2^2+...+2^n=2^{n+1}-1\forall nϵN$

Check that statement is true for $n=1$

We observe that P(n) is true for $n=1$,

Since, $1+2=2^{1+1}-1$

Assume P(k) to be true and then prove $P(k+1)$ is true.

Assume that P(n) is true for $n=k$

$\Rightarrow 1+2+2^2+...+2^k=2^{k+1}-1\dots \left(1\right)$

To prove: $1+2+2^2+....+2^k+1=2^{k+2}-1$

LHS $=1+2+2^2+...+2^k+2^{k+1}$

LHS $=2^{k+1}-1+2^{k+1}$

$={2.2}^{k+1}-1$

$=2^{k+1}+1-1$

$=2^{k+2}-1=RHS$

Thus,$P(k+1)$ is true whenever P(k) is true.

Hence, By Principle of mathematical Induction P(n) is true for all natural numbers n.