Question

1 + 2 + 3 + ... + n = $\frac{n(n+1)}{2}$ i.e. the sum of the first n natural numbers is $\frac{n(n+1)}{2}$.

Answer 1

Let P(n) be the given statement.
Now,
P(n) = 1 + 2 + 3 +...+ n =$\frac{n(n+1)}{2}$
$$\begin{gathered} \mathrm{Step}1: \\ P\left(1\right)=1=\frac{1(1+1)}{2}=1 \\ \mathrm{Hence},P\left(1\right)\mathrm{is}\mathrm{true}. \\ \mathrm{Step}2: \\ \mathrm{Let}P\left(m\right)\mathrm{be}\mathrm{true}. \\ \mathrm{Then}, \\ 1+2+3+...+m=\frac{m(m+1)}{2} \\ \mathrm{We}\mathrm{shall}\mathrm{now}\mathrm{prove}\mathrm{that}P(m+1)\mathrm{is}\mathrm{true}. \\ \mathrm{i}.\mathrm{e}., \\ 1+2+...+(m+1)=\frac{(m+1)(m+2)}{2} \\ \mathrm{Now}, \\ 1+2+...+m=\frac{m(m+1)}{2} \\ \Rightarrow 1+2+...m+m+1=\frac{m(m+1)}{2}+m+1\left[\mathrm{Adding}\left(m+1\right)\mathrm{to}\mathrm{both}\mathrm{sides}\right] \\ =\frac{m^2+m+2m+2}{2} \\ =\frac{(m+1)(m+2)}{2} \\ \mathrm{Hence},P(m+1)\mathrm{is}\mathrm{true}. \end{gathered}$$
By the principle of mathematical induction, P(n) is true for all n$\in$N.