How to solve $\binom{n}{1}^2+2\binom{n}{2}^2 + 3\binom{n}{3}^2 + 4\binom{n}{4}^2+\cdots + n\binom{n}{n}^2$?

Question

I have tried something to solve the series $$\binom{n}{1}^2+2\binom{n}{2}^2 + 3\binom{n}{3}^2 + 4\binom{n}{4}^2+\cdots + n\binom{n}{n}^2.$$ My approach is : $$(1+x)^n=\binom{n}{0} + \binom{n}{1}x + \binom{n}{2}x^2 + \cdots + \binom{n}{n}x^n.$$ Differentiating the above equation $$n(1+x)^{n-1} = \binom{n}{1} + \binom{n}{2}x + \cdots + n\binom{n}{n}x^{n-1}$$

Also, $$ \left(1+\frac{1}{x}\right)^n =\binom{n}{0} + \binom{n}{1}\frac{1}{x} + \binom{n}{2}\left(\frac{1}{x}\right)^2 + \cdots + \binom{n}{n}\left(\frac{1}{x}\right)^n$$ Multiplying above two equation I get, $$\begin{align*} &{n(1+x)^{n-1}\left(1 + \frac{1}{x}\right)^n}\\ &\quad= \left( \binom{n}{1}^2 + 2\binom{n}{2}^2 + 3\binom{n}{3}^2 + 4\binom{n}{4}^2 + \cdots + n\binom{n}{n}^2\right)\left(\frac{1}{x}\right) + \text{other terms} \end{align*}$$

So I can say that coefficient of $\frac{1}{x}$ in expansion of $n(1+x)^{n-1}(1+\frac{1}{x})^n$ will give me the required answer.

Am I doing it correct,please correct me if I'm wrong ?

If I'm right,please tell me how to calculate the coefficient of $\frac{1}{x}$ ?

Based on the answers,I tried to implement the things in a C++ code.

I tried implementing the code using extended euclidean algorithm so that the problem of truncated division can be eliminated but still not abled to figure out why am I getting wrong answer for n>=3. This is my updated code : http://pastebin.com/imS6rdWs I'll be thankful if anyone can help me to figure out what's wrong with this code.

Thanks.

Solution:

Finally abled to solve the problem.Thanks to all those people who spent their precious time for my problem.Thanks a lot.This is my updated code :

http://pastebin.com/WQ9LRy6F

Answer 1

First recall that the coefficient of $x^n$ in $(1+x)^n(1+x)^n=(1+x)^{2n}$ implies $$ \begin{align} \sum_{k=0}^n\binom{n}{k}^2 &=\sum_{k=0}^n\binom{n}{k}\binom{n}{n-k}\\ &=\binom{2n}{n}\tag{1} \end{align} $$ and then note that $$ \begin{align} \sum_{k=0}^nk\binom{n}{k}^2 &=\sum_{k=0}^nk\binom{n}{n-k}^2\\ &=\sum_{k=0}^n(n-k)\binom{n}{k}^2\tag{2} \end{align} $$ Adding the first and last parts of $(2)$ yields $$ \begin{align} 2\sum_{k=0}^nk\binom{n}{k}^2 &=n\sum_{k=0}^n\binom{n}{k}^2\\ &=n\binom{2n}{n}\tag{3} \end{align} $$ Therefore, $$ \sum_{k=0}^nk\binom{n}{k}^2=\frac{n}{2}\binom{2n}{n}\tag{4} $$

Answer 2

This kind of looks like you want to appeal to Vandermonde's convolution, or at least the method you'd use to prove it. It can be applied directly as follows:

Let $S = \sum\limits_{k=0}^n k\binom{n}{k}^2$ be the sum we want to compute. Note that $S = \sum\limits_{k=0}^n (n-k) \binom{n}{n-k}^2 = \sum\limits_{k=0}^n (n-k) \binom{n}{k}^2$. Therefore $2S = n\sum\limits_{k=0}^n \binom{n}{k}^2 = n \binom{2n}{n}$. Then

$$S = \frac{n}{2}\binom{2n}{n}.$$

Answer 3

Building on the excellent work of @thomas-belulovich & @robjohn, but addressing your subsequently revealed goal....

You want $$ T(n) =S(n)-m\left\lfloor\frac{S(n)}m\right\rfloor =S(n)\text{ mod }m \qquad\text{for}\qquad S(n) =\frac{n}{2}{2n\choose n}. $$ Note that $$ \frac{S(n)}{S(n-1)} =\frac{n}{n-1}\cdot\frac{2n(2n-1)}{n^2} =2\frac{2n-1}{n-1}. $$ A brute force approach to this might be to calculate $$ R(k)=2(2k-1)(k-1)^{-1}\pmod m $$ for each $1<k\le n$ and then multiply modulo $m$ termwise to obtain $$ T(n)=S(1)\prod_{k=2}^nR(k). $$ Calculating each $R(k)=(4k-2)x$ requires $O(\log k)$ divisions using the extended Euclidean algorithm to find an $x$ so that $$ (k-1)x+my=1. $$

Extended Euclidean Algorithm: Given nonzero $a,b\in\mathbb{Z}$, find $x,y\in\mathbb{Z}$ so that $ax+by=\text{gcd}(a,b)$. (Adapted from David Joyner's book.)

1. Set $\overline{u}=\langle{u_0,u_1,u_2}\rangle$ $\leftarrow\langle{a,1,0}\rangle$ and $\overline{v}\leftarrow\langle{b,0,1}\rangle$ (vectors in $\mathbb{Z}^3$). Set $\overline{v}\leftarrow-\overline{v}$ if $b<0$.

2. While $v_0 \ne 0$, repeat:

3. $\qquad$ Calculate the quotient $q=\text{quo}(u_0,v_0)=u_0-v_0\left\lfloor\frac{u_0}{v_0}\right\rfloor$.

4. $\qquad$ Set $\quad\overline{w}=\overline{u}-q\overline{v},\quad$ then $\quad\overline{u}=\overline{v},\quad$ and then $\quad\overline{v}=\overline{w}.\quad$

5. Return $a=u_0$, $x=u_1$, and $y=u_2$.

This is doable in not too many lines of C code, and it works as long as $m=10^9+7$ has no (prime) factors $p\le n$ (in fact, this $m$ is prime). If you needed a more efficient algorithm and $m$ were a composite product of distinct primes (which it isn't), you could use the prime factorization of $m$ and a nice fact about binomial coefficients modulo primes [Lukas 1878] to separately calculate residues $$ {a\choose b}\equiv{[a/p]\choose[b/p]}{a\mod p\choose b\mod p}\pmod p $$ modulo each factor $p$, and then recover $T(n)$ using the Chinese Remainder Theorem.

Here's a few pre-computed values: $$\matrix{ k& n=10^k &{2n\choose n}\text{ mod }m &T(n) \\0& 1 &2 &1 \\1& 10 &184756 &923780 \\2& 100 &407336795 &366839610 \\3& 1000 &72475738 &237868748 \\4& 10000 &703593270 &966325381 \\5& 100000 &879467333 &366342189 \\6& 1000000 &192151600 &799327475 }$$

If you want to find an efficient method to solve this problem, you'll probably want to look at the works of Donald Knuth, Andrew Granville, and Richard Stanley, who also has compiled outstanding lists of bijective proof problems and characterizations of the Catalan numbers $C_n=\frac1{n+1}{2n\choose n}$, to which our $S(n)$ are closely related since $S(n)={n+1\choose2}C_n$.

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One might be tempted to try to shorten the computation using Wilson's theorem, $$ p\text{ prime} \quad\implies\quad (p-1)!\equiv-1\pmod p $$ Morley's (1895) congruence, $$ p=2n+1\text{ prime} \quad\implies\quad {2n\choose n}\equiv(-1)^n4^{2n}\pmod{p^3} $$ and/or Kummer's theorem using enough "landmark" primes near $n$ and $2n$, with extended Euclidean algorithm for inverses and the CRT (there's the catch!) to get the final result. For example, here are some prime pairs $q_i=2p_i+1$ near $n=10^6$ and $2n$: $$ \matrix{ i & p_i-n & q_i-2n \\ 1 & -251 & -501 \\ 2 & -191 & -381 \\ 3 & 151 & 303 \\ 4 & 193 & 387 \\ } $$ From Wilson's theorem for odd primes $q$, grouping $(q-1)!=(1\cdots\frac{q-1}2)(\frac{q+1}2\cdots q-1)$ and noting that the second term is $(-1)^\frac{q-1}2$ times the first term modulo $q$, we find that $$ \left(\left(\tfrac{q-1}{2}\right)!\right)^2 \equiv(-1)^{\frac{q+1}2} \pmod q $$ so that for prime pairs $q_i=2p_i+1$, $$ {2p_i\choose p_i}\equiv(-1)^{p_i}=-1\pmod{q_i}. $$ Thus we can compute (using the extended Euclidean algorithm for the inverses of $k$ modulo $q_i$) $$ {2n\choose n}\equiv-\prod_{k=p_i+1}^n\frac{4k-2}{k} \pmod{q_i} $$ for $i=1,2$ above. However, we cannot use the CRT to get $T(n)$. We would have to have enough prime pairs to reconstruct $S(n)$, and then compute its residue at the end. Since the central binomial coefficient is approximately $${2n\choose n}\approx\frac{4^n}{\sqrt{\pi n}}\left(1-\frac1{nc_n}\right),\qquad c_n\in(8,9)$$ we would need about 96 thousand pairs (p,q), making this approach infeasible.

Answer 4

First we address the overflow issue. Note that $10^9+7$ is relatively prime to all the numbers that come up in a naive calculation of $\binom{2n}{n}$. Indeed $10^9+7$ happens to be prime. So when we are calculating, we can always reduce modulo $10^9+7$ as often as necessary to prevent overflow.

Now to the identity. We have $n$ boys and $n$ girls. We want to choose $n$ people. The number of ways this can be done is $\binom{2n}{n}$. But we can choose $0$ boys and $n$ girls, or $1$ boy and $n-1$ girls, and so on. We can choose $k$ boys and $n-k$ girls in $\binom{n}{k}\binom{n}{n-k}$ ways, or equivalently in $\binom{n}{k}^2$ ways. This gives the standard derivation of the identity $$\binom{2n}{n}=\sum_{k=0}^n \binom{n}{k}^2.$$ Note now that the $\binom{2n}{n}$ choices are all equally likely. The expected number of boys is, by symmetry, equal to $\frac{n}{2}.$ But the probability that there are $k$ boys is $\frac{\binom{n}{k}^2}{\binom{2n}{n}}$, and therefore the expected number of boys is $$\sum_{k=0}^n k\frac{\binom{n}{k}^2}{\binom{2n}{n}}.$$ The term corresponding to $k=0$ is $0$, so can be omitted, and we get $$\sum_{k=1}^n k\frac{\binom{n}{k}^2}{\binom{2n}{n}}=\frac{n}{2},$$ which is essentially our identity.

Remark: There is a very nice book on bijective proofs called Proofs that Really Count. A title that so far doesn't seem to have been used is Mean Proofs.