Let P(n):
![\[1\text{ }+\text{ }2\text{ }+\text{ }{{2}^{2}}~+\text{ }\ldots \text{ }+\text{ }{{2}^{n}}~=\text{ }{{2}^{n}}{{~}^{+1}}~\text{ }1\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-39ce23d2f3d91257b994162c1f492d4d_l3.png "Rendered by QuickLaTeX.com")
, for all natural numbers n
![\[P\left( 1 \right):\text{ }1\text{ }={{2}^{0\text{ }+\text{ }1}}~\text{ }1\text{ }=\text{ }2\text{ }\text{ }1\text{ }=\text{ }1,\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-718183a8e87989633ddc1cdadce35e93_l3.png "Rendered by QuickLaTeX.com")
which is true.
Hence, ,P(1) is true.
Let us assume that P(n) is true for some natural number n = k.
![\[P\left( k \right):\text{ }l+2\text{ }+\text{ }{{2}^{2}}+\ldots +{{2}^{k}}~=\text{ }{{2}^{k+1}}-l\text{ }~\text{ }~\text{ }~\text{ }~\text{ }~\text{ }~\text{ }~\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-d7d9434b02d65f578c445f8f9ec010ab_l3.png "Rendered by QuickLaTeX.com")
(i)
Now, we have to prove that P(k + 1) is true.
![\[P\left( k+1 \right):\text{ }1+2\text{ }+\text{ }{{2}^{2}}+\text{ }\ldots +{{2}^{k}}~+\text{ }2k+1\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-5fe2d635fc44c06dddad73f94455814c_l3.png "Rendered by QuickLaTeX.com")
= 2k +1 – 1 + 2k+1 [Using (i)]
![\[=\text{ }{{2.2}^{k+l}}\text{ }1\text{ }=\text{ }1\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-6bab3a4065463bf9dc0afff9585c8ee9_l3.png "Rendered by QuickLaTeX.com")
![\[_{=~}{{2}^{\left( k+1 \right)+1}}-1\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-4f70acd09a88b7aa6cafa3cf42100519_l3.png "Rendered by QuickLaTeX.com")
Hence,
![\[P\left( k\text{ }+\text{ }1 \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-ba7adf40f638ab16e5c43aafa4dab24d_l3.png "Rendered by QuickLaTeX.com")
is true whenever P(k) is true.
So, by the principle of mathematical induction P(n) is true for any natural number n.