Question

Prove that $1+2+3+.....+n=$ $\frac{n(n+1)}{2}$.

Answer 1

Let P(n): 1 + 2 + ..... + n = $\frac{n(n+1)}{2}$ be the given statement.
Step 1: Put n = 1
Then, L.H.S. = 1 and R.H.S. = $\frac{1(1+1)}{2}=1$
$\therefore$ L.H.S = R.H.S.
Step 2: Assume that P(n) is true for n = k.
$\therefore 1+2+3.....+k=\frac{k(k+1)}{2}$
Adding (k + 1) on both sides we get
$1+2+3.....+k+(k+1)=\frac{k(k+1)}{2}$
$=\left(\frac{k}{2}+1\right)$
$=\frac{(k+1)(k+2)}{2}$
$=\frac{(k+1)(\overline{k+1}+1)}{2}$
$\Rightarrow P\left(n\right)$ is true n = k + 1
$\therefore$ By the principle of mathematical induction
p(n) istrue for all natural numbers n.
Hence, 1 + 2 + 3 + ..... + n = $\frac{n(n+1)}{2}$ for all n $ϵ$ N