<s>[INST] Find $X/1430$ when $X=(^{10}C_1)^2+2(^{10}C_2)^2+3(^{10}C_3)^2+ ...+10(^{10}C_{10})^2$ 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.
 [/INST] 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 $x^{10}$.
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