let us suppose that we have following recurrence relation
$a_n-a_{n-1}=2*n$ where $a_0=0$
as i know solution is following
$a_n=a_0+\sum\limits_{i=1}^n 2n $
which is the same as $a_n=2*\sum\limits_{i=1}^n n $
but this term
$ \sum\limits_{i=1}^n n=n^2$
therefore $a_n=2*n^2$
but youtube says that
and also wolfram alpha says that solution is $a_n=n*(n+1)$ how?
The following two summations are correct, \begin{align} \sum_{i=1}^n i &= \frac{n(n+1)}{2} \\ \sum_{i=1}^n n &= n \sum_{i=1}^n 1 = n(n) = n^2 \end{align} For your recurrence relation, we have that at each iteration you extract a term of the form $2i$ as such \begin{equation} \begin{split} a_n &= a_{n-1} + 2n\\ &= a_{n-2} + 2(n-1) + 2n\\ &= \vdots \\ &= a_{i} + 2(i+1) + \ldots + 2n \end{split} \end{equation} For $i = 0$, we get $$a_n = a_0 + \sum_{i=1}^n 2i = 0 + 2 \sum_{i=1}^n i = 2 \frac{n(n+1)}{2}=n(n+1)$$