How to calculate the sum of the first n natural numbers? I already know one way to prove the sum of the n first natural number is equal to $\frac{n(n+1)}{2}$, but I found another way which involves calculating $(k+1)^2 - k^2, k\in \mathbb{N},\ k\geqslant 1$ . Here is the whole demonstration:
$
\begin{gather}
(k+1)^2 -k^2 = 2k +1\\
\sum_{k=1}^n{((k+1)^2 -k^2)} = 2\sum_{k=1}^n{k} +\sum_{n=1}^n{1} \\
(n+1)^2-1=2\sum_{k=1}^n{k} +n \\
2\sum_{k=1}^n{k} = n^2 +n\\
\sum_{k=1}^n{k} = \frac{n(n+1)}{2}
\\
\end{gather}$
My question is: How do we get from $\sum_{k=1}^n{((k+1)^2 -k^2)}$ to $(n+1)^2-1$

Let's try writing some terms out: 
\begin{align*}
\sum\limits_{k=1}^n ((k+1)^2 - k^2) & = ((1+1)^2 - 1^2) + ((2+1)^2 - 2^2) + ((3+1)^2 - 3^2) + \ldots + ((n+1)^2 - n^2) \\
& = (2^2 - 1^2) + (3^2 - 2^2) + (4^2 - 3^2) + \ldots + ((n+1)^2 - n^2) \\
& = (2^2 - 2^2) + (3^2 - 3^2) + \ldots + (n^2 - n^2) + (n+1)^2 - 1^2 \\
& = (n+1)^2 - 1
\end{align*}
It telescopes! 
