# summation of binomial coefficients with squares

What is $$50^2\frac{{n\choose 50}}{{n\choose 49}}+49^2\frac{{n\choose 49}}{{n\choose 48}}...1^2\frac{{n\choose 1}}{{n\choose 0}}$$.

i.e. $$\sum_{k=1}^{50} \frac{k^2\binom n k}{\binom n {k-1}}= ?$$

$$\text{MY ATTEMPT}$$ When I plugged $n=50$ I got $22100$ as the series is wave like, with maximum at $25$ which is $25\times 26$. So I calculated it on a calculator and multiplied the summation till $25\times 26$ and got it. But I want to calculate the general formula for summation. The answer given is $425(3n-98)$. Thanks!

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It's not clear where that first parenthesis is supposed to close, but do you want an alternating sum? – Thomas Andrews Feb 26 at 13:55
If you'd fix your question so that people can figure out what you are asking, people might be able to help. Help people help you. – Thomas Andrews Feb 26 at 14:03

$$\require{cancel} \sum_{r=1}^{50}r^2\frac{\binom nr}{\binom n{r-1}}\color{lightgrey}{=\sum_{r=1}^{50}r^\bcancel2\frac{\frac{\cancel{n(n-1)\cdots (n-r)}(n-r+1)}{\bcancel{r}\cancel{(r-1)!}}}{\frac{\cancel{n(n-1)\cdots (n-r)}}{\cancel{(r-1)!}}}}=\sum_{r=1}^{50}r(n-r+1)\\ =\sum_{r=1}^{50}nr-r(r-1)=\sum_{r=1}^{50} n\binom r1-2\binom {r}2\\ =n\binom {51}2-2\binom {51}3\\ =\frac 1{52}\binom {52}3(3n-98)\\ =425(3n-98)\quad\blacksquare$$

Note that if $n=50$, this reduces to $\binom {52}3$ as can be seen from above. This can also be proven directly as follows:

$$\sum_{r=1}^n r(n-r+1)=\sum_{r=1}^n\sum_{j=r}^nr=\sum_{j=1}^n\sum_{r=1}^j \binom r1=\sum_{j=1}^n\binom {j+1}2=\binom {n+2}3=\binom {52}3$$

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If the sum that you're interested in is: $$\sum_{k=1}^{50}\frac{k^2\binom{n}{k}}{\binom{n}{k-1}},$$ Then, $$\sum_{k=1}^{50}\frac{k^2\binom{n}{k}}{\binom{n}{k-1}}=\sum_{k=1}^{50}\frac{k^2(n-k+1)}{k}=\sum_{k=1}^{50}k(n-k+1)=(n+1)\sum_{k=1}^{50}k-\sum_{k=1}^{50}k^2.$$ Then, using the formulas $$\sum_{k=1}^{50}k=\frac{50\cdot 51}{2}=1275$$ and $$\sum_{k=1}^{50}k^2=\frac{50\cdot 51\cdot 101}{6}=42925,$$ you will get the negative of the answer at the end of the post (perhaps due to the negative confusion in the discussion above).

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which is $1275 n -41650 = 425(3n-98)$ and gives $22100$ when $n=50$. – Henry Feb 26 at 14:18

What to compute is not very clear. Anyway, if you want the value of: $$50^2\frac{\binom{n}{50}}{\binom{n}{49}}+49^2\frac{\binom{n}{49}}{\binom{n}{48}}+\ldots+1^2\frac{\binom{n}{1}}{\binom{n}{0}}=50(n-49)+49(n-48)+\ldots+1(n-0)$$ you just need to recall that $\sum_{k=1}^{m}k = \frac{m(m+1)}{2}$ and $\sum_{k=1}^m k^2 = \frac{m(m+1)(2m+1)}{6}$.

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Hint:

$$\frac{\binom{n}{p}}{\binom{n}{p-1}}=\frac{1}{p(n-p)}$$

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Okay but how to deal with $1.50+49.2+48.3+....+26.25)$ – Archis Welankar Feb 26 at 13:47
Can you check the location of the $n-p+1$ term from the $(n-(p-1))!$? – Michael Burr Feb 26 at 14:11
$\frac{\binom{n}{p}}{\binom{n}{p-1}}=\frac{\frac{n!}{p!(n-p)!}}{\frac{n!}{(p-1)!‌​(n-(p-1))!}}= \frac{(p-1)!(n-(p-1))!}{p!(n-p)!}= \frac{(p-1)!(n-p+1)!}{p!(n-p)!}= \frac{n-p+1}{p}$. – Michael Burr Feb 26 at 14:21
perhaps you meant $\frac {\frac np}{\frac n{p-1}}=\frac 1{p(n-p)}$... – hypergeometric Feb 26 at 14:38
I was referring to @Bacon's answer and trying to point that out. – hypergeometric Feb 26 at 14:45

Each term has the form

$$k^2\frac{\binom{n}{k}}{\binom{n}{k-1}}$$

We can simplify this

$$k^2\frac{\binom{n}{k}}{\binom{n}{k-1}} = \frac{k^2\frac{n!}{(n-k)!k!}}{\frac{n!}{(n-k+1)!(k-1)!}} = \frac{k^2*n!(n-k+1)!(k-1)!}{n!*k!(n-k)!} = \frac{k^2(n-k+1)}{k} = k(n-k+1)$$

Thus

\begin{align} \sum_{k=1}^{\hat{K}} k^2\frac{\binom{n}{k}}{\binom{n}{k-1}} &=& \sum_{k=1}^{\hat{K}} k(n-k+1) \\&=& (n+1)\sum_{k=1}^{\hat{K}} k - \sum_{k=1}^{\hat{K}} k^2 \\&=& \frac{\hat{K}(n+1)(\hat{K}+1)}{2} - \frac{\hat{K}(\hat{K}+1)(2\hat{K}+1)}{6} \\&=& \frac{\hat{K}(3n - 2\hat{K} + 2)(\hat{K}+1)}{6} \end{align}

Now, in your specific case, $\hat{K} = 50$, so we get a sum of $425(3n-98)$ by plugging this into the above.

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