Question

Find the sum of the first n natural numbers.

• A. $\frac{n(n+1)}{2}$
• B. $\frac{(n+1)}{2}$
• C. $\frac{(n-1)(n+1)}{2}$
• D. $\frac{n(n-1)}{2}$

Answer 1

The correct option is A

$\frac{n(n+1)}{2}$

First n natural numbers 1,2,3,4,5,6,.................... n terms

$S_n$ = 1+2+3+4+5+6+..........(n-2) + (n-1) + n ------------(1)

We can also write above given series in reverse order

$S_n$ = n + (n-1) + (n-2) +...............4,3,2,1 ----------------(2)

we find sum of first and last term = sum of second and second last term so,on.....

Equation 1 + Equation 2

2$S_n$ = (n+1) + (n+1) + (n+1) ......... n times

2$S_n$ =n.(n+1)

$S_n$ = $\frac{n(n+1)}{2}$