Proving that n²+n is always even

Prove that if n is an integer then n²+n is even.

Let n be an integer. Then It is either odd or even. Suppose that n is even. Then n=2k for some integer k and in this case,

n²+n=4k²+2k=2(2k²+k)=2ℓ

where ℓ is an integer defined as ℓ=2k²+k. Suppose next that n is odd. Then n=2k+1 for some integer k and in this case,

n²+n  = 4(2k+1)²+2(2k+1)
 = 16k²+20k+6
 = 2(8k²+10k+3)
 = 2ℓ.

where ℓ is an integer defined as ℓ=8k²+10k+3. In conclusion, the parity of n in inconsequential when investigating the parity of n²+n. In both cases, we have indeed established that n²+n is always even.