Question

Prove that $n^2+n$ is divisible by $2$ for any positive integer $n$.

Answer 1

Let $m=n^2+n=n(n+1)$

Clearly seen, the $m$ is product of two consecutive

integers. i.e. our even and one odd

$\therefore$ when dividing from ${}^{\prime }{2}^{\prime }$ it will be divisible because

there is always an even team preset.

$\frac{No}{2}=\frac{n(n+1)}{2}=\sum$ of integers from $0$ to $\infty$