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If $C_r$ denotes the binomial coefficient ${ }^n C_r$, then $(-1) C_0^2+2 C_1^2+5 C_2^2+\ldots+(3 n-1) C_n^2$ is equal to
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Verified Answer
The correct answer is:
$\left(\frac{3 n-2}{2}\right){ }^{2 \pi} C_n$
Consider the given expression,
$$
\begin{aligned}
& (-1) C_0^2+2 C_1^2+5 C_2^2+\ldots+(3 n-1) C_n^2 \\
& =\sum_{r=0}^n(3 r-1) C_r^2 \\
& =\sum_{r=0}^n 3 r \cdot C_r^2-\sum_{r=0}^n C_r^2 \\
& =3 \sum_{r=0}^n\left(r \cdot C_r^2\right)-{ }^{2 n} C_n \\
& =3 \sum_{r=0}^n\left(r \cdot C_r^2\right)-{ }^{2 n} C_n \\
& =3 \sum_{r=1}^n\left(r \cdot{ }^n C_r \cdot{ }^n C_r\right)-{ }^{2 n} C_n \\
& \left.=3 \sum_{r=1}^n\left(n \cdot{ }^{n-1} C_{r-1}^2 \cdot{ }^n C_r\right)-{ }^{2 n} C_n+\ldots+C_n^2={ }^{2 n} C_n\right]
\end{aligned}
$$
$\begin{aligned} & \quad=3 n \sum_{r=0}^n\left({ }^{n-1} C_{r-1}{ }^n C_{n-r}\right)-{ }^{2 n} C_n \\ & \qquad\left[\because r \cdot{ }^n C_r=n \cdot{ }^{n-1} C_{r-1}\right] \\ & =3 n\left[\text { coefficient of } x^{n-1} \text { in the expansion of }\right. \\ & \left.\qquad(1+x)^{n-1}(1+x)^n\right]-{ }^{2 n} C_n \\ & =3 n\left[\text { coefficient of } x^{n-1} \text { in the expansion of }\right. \\ & \left.\quad(1+x)^{2 n-1}\right]-{ }^{2 n} C_n \\ & =3 n{ }^{2 n-1} C_{n-1}-{ }^{2 n} C_n \\ & =3 n \frac{(2 n-1) !}{(n-1) !(n) !}-\frac{(2 n) !}{(n !)(n !)}\end{aligned}$
$$
\begin{aligned}
& =\frac{3 n \cdot(2 n)(2 n-1) !}{(2 n)(n-1) !(n !)}-\frac{(2 n) !}{n ! n !}=\frac{3 n}{2} \frac{(2 n) !}{n ! n !}-\frac{(2 n) !}{n ! n !} \\
& =\frac{3 n}{2} \cdot{ }^{2 n} C_n-{ }^{2 n} C_n={ }^{2 n} C_n\left(\frac{3 n}{2}-1\right) \\
& =\left(\frac{3 n-2}{2}\right) \cdot{ }^{2 n} C_n
\end{aligned}
$$
$$
\begin{aligned}
& (-1) C_0^2+2 C_1^2+5 C_2^2+\ldots+(3 n-1) C_n^2 \\
& =\sum_{r=0}^n(3 r-1) C_r^2 \\
& =\sum_{r=0}^n 3 r \cdot C_r^2-\sum_{r=0}^n C_r^2 \\
& =3 \sum_{r=0}^n\left(r \cdot C_r^2\right)-{ }^{2 n} C_n \\
& =3 \sum_{r=0}^n\left(r \cdot C_r^2\right)-{ }^{2 n} C_n \\
& =3 \sum_{r=1}^n\left(r \cdot{ }^n C_r \cdot{ }^n C_r\right)-{ }^{2 n} C_n \\
& \left.=3 \sum_{r=1}^n\left(n \cdot{ }^{n-1} C_{r-1}^2 \cdot{ }^n C_r\right)-{ }^{2 n} C_n+\ldots+C_n^2={ }^{2 n} C_n\right]
\end{aligned}
$$
$\begin{aligned} & \quad=3 n \sum_{r=0}^n\left({ }^{n-1} C_{r-1}{ }^n C_{n-r}\right)-{ }^{2 n} C_n \\ & \qquad\left[\because r \cdot{ }^n C_r=n \cdot{ }^{n-1} C_{r-1}\right] \\ & =3 n\left[\text { coefficient of } x^{n-1} \text { in the expansion of }\right. \\ & \left.\qquad(1+x)^{n-1}(1+x)^n\right]-{ }^{2 n} C_n \\ & =3 n\left[\text { coefficient of } x^{n-1} \text { in the expansion of }\right. \\ & \left.\quad(1+x)^{2 n-1}\right]-{ }^{2 n} C_n \\ & =3 n{ }^{2 n-1} C_{n-1}-{ }^{2 n} C_n \\ & =3 n \frac{(2 n-1) !}{(n-1) !(n) !}-\frac{(2 n) !}{(n !)(n !)}\end{aligned}$
$$
\begin{aligned}
& =\frac{3 n \cdot(2 n)(2 n-1) !}{(2 n)(n-1) !(n !)}-\frac{(2 n) !}{n ! n !}=\frac{3 n}{2} \frac{(2 n) !}{n ! n !}-\frac{(2 n) !}{n ! n !} \\
& =\frac{3 n}{2} \cdot{ }^{2 n} C_n-{ }^{2 n} C_n={ }^{2 n} C_n\left(\frac{3 n}{2}-1\right) \\
& =\left(\frac{3 n-2}{2}\right) \cdot{ }^{2 n} C_n
\end{aligned}
$$
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