Question

Prove that for any integer n, n(n+1)/2 is also an integer.

Answer 1

To prove n(n+1)/2 is also an integer, we need to prove n(n+1) is divisible by 2.
We can prove it case by case.
Case 1: n is divisible by 2, then we can say n=2m, where m is an integer.
n(n+1) / 2 = 2m(2m+1) / 2 = m(2m+1), which is an integer.
Case 2: n is not divisible by 2, then we can say that n=2m+1, where m is an integer. In this case,
n(n+1) / 2 = (2m+1)[(2m+1)+1] / 2 = (2m+1)(2m+2)/2 = ...
I think you can work it out from here :)

Answer 2

A simple way to understand the concept behind the above answer is:
1. N and (N+1) are consecutive integers. Therefore one of them must be Even and the other Odd.
2. Thus, in the product N.(N+1), one of the number N or (N+1) is even and so divisible by 2.
3. Hence the product N.(N+1) itself is divisible by 2 and thus an integer itself.

Answer 3

You could also recall that
\[\large\sum_{k=1}^nn=\frac{n(n+1)}{2}\]
The series is a summation of natural numbers (integers), so the sum must also be a natural number (integer).

Answer 4

Oops, should be
\[\large\sum_{k=1}^n\color{red}k=\frac{n(n+1)}{2}\]