Let P(n):\[1\text{ }+\text{ }2\text{ }+\text{ }{{2}^{2}}~+\text{ }\ldots \text{ }+\text{ }{{2}^{n}}~=\text{ }{{2}^{n}}{{~}^{+1}}~\text{ }1\] , for all natural numbers n
\[P\left( 1 \right):\text{ }1\text{ }={{2}^{0\text{ }+\text{ }1}}~\text{ }1\text{ }=\text{ }2\text{ }\text{ }1\text{ }=\text{ }1,\] which is true.
Hence, ,P(1) is true.
Let us assume that P(n) is true for some natural number n = k.

\[P\left( k \right):\text{ }l+2\text{ }+\text{ }{{2}^{2}}+\ldots +{{2}^{k}}~=\text{ }{{2}^{k+1}}-l\text{ }~\text{ }~\text{ }~\text{ }~\text{ }~\text{ }~\text{ }~\]  (i)

Now, we have to prove that P(k + 1) is true.

\[P\left( k+1 \right):\text{ }1+2\text{ }+\text{ }{{2}^{2}}+\text{ }\ldots +{{2}^{k}}~+\text{ }2k+1\] = 2^{k +1} – 1 + 2^k+1  [Using (i)]
_{\[=\text{ }{{2.2}^{k+l}}\text{ }1\text{ }=\text{ }1\]}

\[_{=~}{{2}^{\left( k+1 \right)+1}}-1\]
Hence, \[P\left( k\text{ }+\text{ }1 \right)\] is true whenever P(k) is true.
So, by the principle of mathematical induction P(n) is true for any natural number n.