Question

Prove by induction that for $n\in N,n^2+n$ is an even integer $(n\ge 1)$

Answer 1

$n=1,n^2+n=2$ is an even integer
Let for $n=k,k^2+k$ is even
Now for $n=k+1$.
$(k+1{)}^2+(k+1)-(k^2+k)$
$=k^2+2k+1+k+1-k^2-k=2k+2$
Which is even integer also $k^2+k$ is integer
Hence $(k+1{)}^2+(k+1)$ is also an even integer.
Hence $n^2+n$ is even integer for all $n\in N$.