Let $X=(^{10}C_1)^2+2(^{10}C_2)^2+3(^{10}C_3)^2+ ...+10(^{10}C_{10})^2$, then what's the value of $X\over1430$?
I don't even know where to begin on this question. All solutions I've seen on various sites start by writing this as a summation and simplifying, and eventually bring it into this form:
$$=10\sum_{r=1}^{10}C^{10}_{r}*^{9}C_{r-1}$$. Until this point, I understand, but after this, I dont understand at all.
On
First of all remember that when $n \geq k\geq0:$ $$\binom{n}{k}=\binom{n}{n-k}$$
Now, using Vandermonde identity realize that : $$\sum_{k=0}^{n}\binom{n}{k}^2=\sum_{k=0}^{n}\binom{n}{k}\binom{n}{n-k}=\binom{2n}{n}$$
Moreover , $$\sum_{k=0}^{n}k\binom{n}{k}^2=\sum_{k=0}^{n}k\binom{n}{n-k}^2$$ $$\sum_{k=0}^{n}k\binom{n}{k}^2=\sum_{k=0}^{n}(n-k)\binom{n}{k}^2$$
By summing these two foregoing identity :
$$2\sum_{k=0}^{n}k\binom{n}{k}^2=n\sum_{k=0}^{n}\binom{n}{k}^2$$
The right hand side says that use Vandermodes' identity , so : $$n\sum_{k=0}^{n}\binom{n}{k}^2=n\binom{2n}{n}$$ Then ,$$2\sum_{k=0}^{n}k\binom{n}{k}^2=n\binom{2n}{n}$$ $$\sum_{k=0}^{n}k\binom{n}{k}^2=\bigg(\frac{n}{2}\bigg)\binom{2n}{n}$$
In this question , we have $n=10$ ,so $$X=5\binom{20}{10}=5\times184,756=923,780$$ $$\frac{X}{1430}=\frac{923,780}{1430}=646$$
On
Here is a variation using the binomial theorem and a bit of algebra. We use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ in a series. This way we can write for instance \begin{align*} \binom{r}{k}=[x^k](1+x)^r\tag{1} \end{align*}
We obtain \begin{align*} \color{blue}{X}&\color{blue}{=\sum_{r=1}^{10}r\binom{10}{r}^2}=10\sum_{r=1}\binom{10}{r}\binom{9}{r-1}\tag{2}\\ &=10\sum_{r=1}^{10}\binom{10}{r}\binom{9}{10-r}\tag{3}\\ &=10\sum_{r=1}^{10}\binom{10}{r}[x^{10-r}](1+x)^9\tag{4}\\ &=10[x^{10}](1+x)^9\sum_{r=1}^{10}\binom{10}{r}x^r\tag{5}\\ &=10[x^{10}](1+x)^9\left((1+x)^{10}-1\right)\tag{6}\\ &=10[x^{10}](1+x)^{19}\tag{7}\\ &\;\;\color{blue}{=10\binom{19}{10}}\tag{8} \end{align*} We finally calculate $\color{blue}{X}=\frac{10}{1\,430}\binom{19}{10}\color{blue}{=646}$.
Comment:
In (2) we use the binomial identity $\binom{r}{k}=\frac{r}{k}\binom{r-1}{k-1}$.
In (3) we use the binomial identity $\binom{r}{k}=\binom{r}{r-k}$.
In (4) we apply the coefficient of operator according to (1).
In (5) we apply the rule $[x^{p-q}]A(x)=[x^p]x^qA(x)$ and factor out terms which do not depend on the summation index $r$.
In (6) we apply the binomial theorem.
In (7) we observe $[x^{10}](1+x)^{9}=0$.
in (8) we select the coefficient of $x^{10}$.
Consider the identity: $$ \begin{aligned} &1+C_1^{19}x+\dots+C_9^{19}x^9+C_{10}^{19}x^{10}+\dots +x^{19} \\ &\qquad=(1+x)^{19} \\ &\qquad=(1+x)^{10}(1+x)^{9} \\ &\qquad= (C_0^{10} + C_1^{10}x+ C_2^{10}x^2+\dots + C_{10}^{10}x^{10}) \cdot (C_0^9 + C_1^9x+ C_2^9x^2+\dots + C_9^9x^9) \ . \end{aligned} $$ Now let us isolate in the start and in the end expression the term in degree ten. $$ \begin{aligned} C_{10}^{19} &= C_{10}^{10}\cdot C_0^9 + C_9^{10}\cdot C_1^9 + C_8^{10}\cdot C_2^9 + C_7^{10}\cdot C_3^9 + C_6^{10}\cdot C_4^9 \\ &\qquad\qquad + C_5^{10}\cdot C_5^9 + C_4^{10}\cdot C_6^9 + C_3^{10}\cdot C_7^9 + C_2^{10}\cdot C_8^9 + C_1^{10}\cdot C_9^9 \ . \end{aligned} $$ The last sum also appears in the question. So we have to compute ten times of it to get $X$, then finally: $$ \begin{aligned} \frac X{1430} &=\frac{10\cdot C_{10}^{19}}{1430} =\frac1{143}\cdot C_{10}^{19} = \frac1{11\cdot 13} \cdot \frac {11\cdot 12\cdot 13\cdot 14\cdot 15\cdot 16\cdot 17\cdot 18\cdot 19} {1\cdot 2\cdot 3\cdot 4\cdot 5\cdot 6\cdot 7\cdot 8\cdot 9} \\ &= \frac {12\cdot 14\cdot 15\cdot 16\cdot 17\cdot 18\cdot 19} {1\cdot 2\cdot 3\cdot 4\cdot 5\cdot 6\cdot 7\cdot 8\cdot 9} = \frac {14\cdot 16\cdot 17\cdot 18\cdot 19} { 4\cdot 7\cdot 8\cdot 9} = \frac {16\cdot 17\cdot 19} { 8} \\ &=2\cdot 17\cdot 19=646\ . \end{aligned} $$ Computer check: