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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 $(n+1)^2-1$

Corentin F
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2 Answers2

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That expression you have, $\sum_{k=1}^n{((k+1)^2 -k^2)}$, is a telescoping series. That means you are adding one term but subtracting previous one each step of the sum, to be said, you have a sum of the form $$\sum_{k=1}^n{(a_{n+1} - a_{n})},$$ with $n\in \mathbb{N}$. That means that the result of that sum will be the last one ($k=n$) minus the first one ($k=1$), so $$\sum_{k=1}^n{(a_{n+1} - a_{n})} = a_{n+1} - a_{1}.$$ So we get that $$\sum_{k=1}^n{((k+1)^2 -k^2)} = (n+1)^2-1.$$

Alejandro Bergasa Alonso
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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!

John P.
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