Evaluate $\sum_{k=0}^nk(n-k)$
Evaluate $\sum_{k=0}^nk(n-k)$
I have $\sum_{k=0}^nkn-\sum_{k=0}^nk^2$
then im suppose to use the identity of $\left( \begin{array}{c} r \\ r \end{array} \right)+\left( \begin{array}{c} r+1 \\ r \end{array} \right)+...+\left( \begin{array}{c} n \\ r \end{array} \right)=\left( \begin{array}{c} n+1 \\ r+1 \end{array} \right)$ to solve this , but i have no idea how this works .
can someone explain in detail of how binomial identities substitute into this ?
Thanks in advance!
$\endgroup$2 Answers
$\begingroup$I hadn’t thought of using that identity, but it works quite nicely:
$$\begin{align*} \sum_{k=0}^nkn&=n\sum_{k=0}^nk\\\\ &=n\sum_{k=0}^n\binom{k}1\;, \end{align*}$$
and
$$\begin{align*} \sum_{k=0}^nk^2&=\sum_{k=0}^n\Big(k(k-1)+k\Big)\\\\ &=\sum_{k=0}^nk(k-1)+\sum_{k=0}^nk\\\\ &=2\sum_{k=0}^n\binom{k}2+\sum_{k=0}^n\binom{k}1\;.\tag{1} \end{align*}$$
I’ll leave the rest to you.
Added: You’ve been given the identity $$\binom{r}r+\binom{r+1}r+\ldots+\binom{n}r=\binom{n+1}{r+1}\;,$$ or in summation notation $$\sum_{k=r}^n\binom{k}r=\binom{n+1}{r+1}\;.$$ Since $\binom{k}r=0$ when $0\le k<r$, this can just as well be written $$\sum_{k=0}^n\binom{k}r=\binom{n+1}{r+1}$$ or, in expanded form, $$\binom0r+\binom1r+\ldots+\binom{n}r=\binom{n+1}{r+1}\;.$$ $(1)$ has two summations of this form, one with $r=2$ and one with $r=1$. For the former, for instance, the identity says that $$\sum_{k=0}^n\binom{k}2=\binom{n+1}{2+1}=\binom{n+1}3\;.$$ If you want, you can convert this to a polynomial in $n$:
$$\begin{align*} \binom{n+1}3&=\frac{(n+1)!}{3!\big((n+1)-3\big)!}\\ &=\frac{(n+1)!}{3!(n-2)!}\\ &=\frac{(n+1)n(n-1)}{3!}\\ &=\frac16n(n-1)(n+1)\\ &=\frac16\left(n^3-n\right)\;. \end{align*}$$
$\endgroup$ 2 $\begingroup$If you don't mind using a different approach then you can do the following.$\sum_{k=0}^n k(n-k)=\sum_{k=0}^n(kn-k^2)=n\sum_{k=0}^n k-\sum_{k=0}^n k^2$.
Now let's use that:$\sum_{k=0}^n k={0+n\over 2}(n+1), \sum_{k=0}^n k^2={n(n+1)(2n+1)\over 6}$.
We get: $n\cdot{n\over 2}(n+1)-{n(n+1)(2n+1)\over 6}={n^3-n\over 6}$.
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