I already know one way to prove the sum of the n first natural number is equal to n(n+1)2n(n+1)2, but I found another way which involves calculating (k+1)2−k2,k∈N, k⩾1 . Here is the whole demonstration:
(k+1)2−k2=2k+1n∑k=1((k+1)2−k2)=2n∑k=1k+n∑n=11(n+1)2−1=2n∑k=1k+n2n∑k=1k=n2+nn∑k=1k=n(n+1)2
My question is: How do we get from ∑nk=1((k+1)2−k2) to (n+1)2−1
