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If $C_j={ }^n C_j$, then $C_0 C_r+C_1 C_{r+1}+C_2 C_{r+2}+\ldots+C_{n-r} C_n=$
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Verified Answer
The correct answer is:
$\frac{(2 n) !}{(n-r) !(n+r) !}$
$\begin{aligned}
& \text {Since }(1+x)^{2 n}=(1+x)^n(x+1)^n \\
& \Rightarrow(1+x)^{2 n}=\left(C_0+C_1 x+C_2 x^2+\ldots+C_n x^n\right) \\
& \left(C_0 x^n+C_1 x^{n-1}+C_2 x^{n-2}+\ldots+C_n\right)
\end{aligned}$
Now equation co-efficient of $x^{n-r}$ on both side
$\Rightarrow C_0 C_r+C_1 C_{r+1}+C_2 C_{r+2}+\ldots+C_{n-r} C_n={ }^{2 n} C_{n+r}$
& \text {Since }(1+x)^{2 n}=(1+x)^n(x+1)^n \\
& \Rightarrow(1+x)^{2 n}=\left(C_0+C_1 x+C_2 x^2+\ldots+C_n x^n\right) \\
& \left(C_0 x^n+C_1 x^{n-1}+C_2 x^{n-2}+\ldots+C_n\right)
\end{aligned}$
Now equation co-efficient of $x^{n-r}$ on both side
$\Rightarrow C_0 C_r+C_1 C_{r+1}+C_2 C_{r+2}+\ldots+C_{n-r} C_n={ }^{2 n} C_{n+r}$
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