Binomial theorem coefficient question Use the binomial theorem to find the coefficient for $x^3$ on both sides of the expansion of:
$(1+x)^3$$(1+x)^3$ $=$ $(1+x)^6$
i. Hence show $(_3C_0)^2+(_3C_1)^2+(_3C_2)^2+(_3C_3)^2$ $=$ $_6C_3$
ii. Use the same argument with $(1+x)^n(1+x)^n = (1+x)^{2n}$ to prove
$\sum_{k=0}^n(_nC_k)^2=_{2n}C_n$
Thank you.
 A: It is convenient to use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$ in a series. We also use the notation $\binom{n}{k}$ instead of $_nC_k$.

We obtain
  \begin{align*}
[x^3](1+x)^3(1+x)^3&=[x^3]\left(\sum_{j=0}^3\binom{3}{j}x^j\right)\left(\sum_{k=0}^3\binom{3}{k}x^k\right)\tag{1}\\
&=\sum_{j=0}^3\binom{3}{j}[x^{3-j}]\sum_{k=0}^3\binom{3}{k}x^k\tag{2}\\
&=\sum_{j=0}^3\binom{3}{j}\binom{3}{3-j}\tag{3}\\
&=\sum_{j=0}^3\binom{3}{j}^2\tag{4}\\
\end{align*}
on the other hand we obtain
  \begin{align*}
[x^3](1+x)^3(1+x)^3&=[x^3](1+x)^6\\
&=[x^3]\sum_{j=0}^6\binom{6}{j}x^j\tag{5}\\
&=\binom{6}{3}
\end{align*}

Comment:


*

*In (1) we apply the binomial theorem twice.

*In (2) we use the linearity of the coefficient of operator and apply the rule
\begin{align*}
[x^p]x^qA(x)=[x^{p-q}]A(x)
\end{align*}
to the summands of the left series.

*In (3) we select the coefficient of $x^{3-j}$ from the right series.

*In (4) we use the symmetry $\binom{n}{k}=\binom{n}{n-k}$.

*In (5) we again apply the binomial theorem and select the coefficient of $x^3$.

We conclude
  \begin{align*}
\sum_{j=0}^3\binom{3}{j}^2=\binom{6}{3}\\
\end{align*}

The calculation to show
\begin{align*}
\sum_{j=0}^n\binom{n}{j}^2=\binom{2n}{n}\\
\end{align*}
can be done analogously.
