Prove the following by using iduction for all `n in N`. `1+2+3+.....+n=(n(n+1))/(2)`

Explanation

Let `P(n):1+2+3+....+n(n(n+1))/(2)=1`
Step I For `n=1`,
LHS of Eq. (i) =1
RHS of Eq. 9i) `=(1(1+1))/(2)=1`
LHS = RHS
Therefore , P(1) is true.
Step II Let us assume that the result is true for `n=k` , Then ,
P(k) `: 1 + 2 + 3 + ... + k = (k(k+ 1))/(2)`
Step III For n = k + 1 , we have to prove that
P(k+ 1) = ` 1 + 2 + 3 + ... + k + (k + 1) = ((k + 1) (k + 2))/(2)`
L.H.S = 1 + 2 + 3 + ... + k+ ( k + 1)
` = (k(k + 1))/(2) + k + 1 ` [ By assumption step]
`= (k + 1) ((k)/(2) + 1) = (k + 1) ((k+ 2)/(2))`
` = (( k + 1 ) (k + 2))/(2)`
= RHS
This show that the result is true for n = k + 1 . Therefore , by the principle of mathematical induction , the result is true for all `n in N`