Finding a formula for $\sum_{k=1}^n(k^2-(k-1)^2)$

Question

I have got this following series: $$\sum_{k=1}^n(k^2-(k-1)^2)$$ I want to come up with a formula for the summation.

I did some math and for me, the formula would be as follows: $$\sum_{k=1}^n(k^2-(k-1)^2)=\sum_{k=1}^n(2k-1)$$ $$=2\sum_{k=1}^nk-\sum_{k=1}^n1=n^2$$ Is that correct? Is the formula for the summation really $n^2$?

Thanks in advance

Answer 1

Yep! If you imagine

$$\sum_{k=1}^n \Bigl(k^2 - (k-1)^2\Bigr), $$

you have the first few terms:

$$(\color{red}{1^2} - 0^2) + (\color{magenta}{2^2} - \color{red}{1^2}) + (\color{green}{3^2} - \color{magenta}{2^2}) +\ldots + \bigl(\color{blue}{(n-1)^2} - \color{orange}{(n-2)^2}\bigr) + \bigl(n^2 - \color{blue}{(n-1)^2}\bigr).$$

The second part of each term cancels out with the first part of the next term up until the last, where there is no next term to cancel out $n^2$. Thus, this series telescopes.

Answer 2

You can have it much easier when you see that this is a telescoping sum

$$\sum_{k=1}^n(a_k - a_{k-1}) = a_n - a_0$$

with $a_k = k^2$.

So, the result is $n^2$.