Question

prove that 1+2+3+.........+n=(n+1)÷2

Answer 1

1+2+3+...+n is an Arithmetic progression with common difference of 1.

so we can find the sum by formula....
Sn =n/2[a+l]...

where l is the last term of the AP.

so summation of first n natural numbers...
Sn=n/2[1+n]=n(n+1)/2