Prove that 1 + 2 + 2

Solution:

Let $\mathrm{p}(n): 1+2+2^{2}+\ldots+2^{n}=2^{n+1}-1 \forall n \in \mathbf{N}$

Step I: For $n=1$,

$\mathrm{LHS}=1+2^{1}=3$

$\mathrm{RHS}=2^{1+1}-1=2^{2}-1=4-1=3$

As, $\mathrm{LHS}=\mathrm{RHS}$

So, it is true for $n=1$.

Step II : For $n=k$,

Let $\mathrm{p}(k): 1+2+2^{2}+\ldots+2^{k}=2^{k+1}-1$ be true $\forall k \in \mathbf{N}$

$\mathrm{LHS}=1+2+2^{2}+\ldots+2^{k}+2^{k+1}$

$=2^{k+1}-1+2^{k+1} \quad($ Using step II $)$

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

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

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

$\mathrm{RHS}=2^{(k+1)+1}-1=2^{k+2}-1$

As, LHS = RHS

So, it is also true for $n=k+1$.

Hence, $1+2+2^{2}+\ldots+2^{n}=2^{n+1}-1$ for all $n \in \mathbf{N}$.