So I have just started learning about summations, and I am stuck with this problem for a while, I need to simplify this summation to get a simpler expression, $\sum_{i=1}^n i(i+1)$ where n is given.
Any help would be much appreciated and Thank you.
How can simplify this summation notation
summation
Related Solutions
You could calculate as follows
\begin{align*} 2\sum_{n=0}^\infty& a_nx^{n+1} +\sum_{n=1}^ \infty nb_nx^{n-1}\\ &=2\sum_{n=1}^\infty a_{n-1}x^{n} +\sum_{n=0}^ \infty (n+1)b_{n+1}x^{n}\tag{1}\\ &=2\sum_{n=1}^\infty a_{n-1}x^{n} +\left(b_1+\sum_{n=1}^ \infty (n+1)b_{n+1}x^{n}\right)\tag{2}\\ &=b_1+\sum_{n=1}^{\infty}\left(2a_{n-1}+(n+1)b_{n+1}\right)x^n\tag{3} \end{align*}
Comment:
In (1) we shift the index of both series by one in order to obtain summands with $x^n$.
In (2) we separate the first summand of the second series. So we get the same index starting value $n=1$ in both series.
In (3) we collect the terms into one series.
If we denote the series (3) with \begin{align*} C(x)=\sum_{n=0}^\infty c_nx^n \end{align*} we observe \begin{align*} c_0&=b_1\\ c_n&=2a_{n-1}+(n+1)b_{n+1}\qquad\qquad n\geq 1 \end{align*}
First, let's prove that for any sequences $(a_{k})$ and $(b_{k})$ and a positive integer $n$, $$ \sum_{k=1}^{n}\left[a_{k}+b_{k}\right]=\sum_{k=1}^{n}a_{k}+\sum_{k=1}^{n}b_{k}. $$ Under the convention $\sum_{k=1}^{0}f_{k}=0$, the statement above is trivially true for $n=0$. Now, suppose the statement holds for some particular $n=n_{0}-1$ with $n_{0}\geq1$. Then, $$ \sum_{k=1}^{n_{0}}\left[a_{k}+b_{k}\right]=a_{n_{0}}+b_{n_{0}}+\sum_{k=1}^{n_{0}-1}\left[a_{k}+b_{k}\right]=a_{n_{0}}+b_{n_{0}}+\sum_{k=1}^{n_{0}-1}a_{k}+\sum_{k=1}^{n_{0}-1}b_{k}=\sum_{k=1}^{n_{0}}a_{k}+\sum_{k=1}^{n_{0}}b_{k}. $$ The desired result follows by induction.
Now, you should be able to use the above result to prove that for any sequences $(a_{k\ell})$ and $(b_{k\ell})$ and positive integers $n$ and $m$, $$ \sum_{k=1}^{n}\sum_{\ell=1}^{m}\left[a_{k\ell}+b_{k\ell}\right]=\sum_{k=1}^{n}\sum_{\ell=1}^{m}a_{k\ell}+\sum_{k=1}^{n}\sum_{\ell=1}^{m}b_{k\ell}. $$
Best Answer
Note that often groups like $i(i+1)$ call for telescopic summation.
It is well known that $\dfrac 1{i(i+1)}=\dfrac 1i-\dfrac 1{i+1}$
Here it is slightly different, we have to increase the degree:
$U_{i}-U_{i-1}=i(i+1)(i+2)-(i-1)i(i+1)=3i^2+3i=3i(i+1)$
Therefore $\sum\limits_{i=1}^n i(i+1)=\dfrac 13(U_n-U_0)=\dfrac 13n(n+1)(n+2)-0$
This works similarly for $\sum i(i+1)(i+2)$ if you want to have a try.