If C_r denotes the binomial coefficient { }^{ | vedclass

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(B) Consider the given expression: $S = \sum_{r=0}^n (3r-1) C_r^2$.We know that $C_r^2 = C_r \cdot C_{n-r}$.So,$S = 3 \sum_{r=0}^n r C_r^2 - \sum_{r=0

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$\left(\frac{3 n-2}{2}\right){ }^{2 n} C_n$

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(B) Consider the given expression: $S = \sum_{r=0}^n (3r-1) C_r^2$.We know that $C_r^2 = C_r \cdot C_{n-r}$.So,$S = 3 \sum_{r=0}^n r C_r^2 - \sum_{r=0}^n C_r^2$.Using the identity $\sum_{r=0}^n C_r^2 = { }^{2n} C_n$ and $r C_r = n { }^{n-1} C_{r-1}$,we have:$S = 3 \sum_{r=1}^n n { }^{n-1} C_{r-1} C_r - { }^{2n} C_n$.$S = 3n \sum_{r=1}^n { }^{n-1} C_{r-1} { }^{n} C_{n-r} - { }^{2n} C_n$.The summation $\sum_{r=1}^n { }^{n-1} C_{r-1} { }^{n} C_{n-r}$ represents the coefficient of $x^{n-1}$ in the expansion of $(1+x)^{n-1}(1+x)^n = (1+x)^{2n-1}$,which is ${ }^{2n-1} C_{n-1}$.Thus,$S = 3n { }^{2n-1} C_{n-1} - { }^{2n} C_n$.Since ${ }^{2n-1} C_{n-1} = \frac{n}{2n} { }^{2n} C_n = \frac{1}{2} { }^{2n} C_n$,we get:$S = 3n \left( \frac{1}{2} { }^{2n} C_n \right) - { }^{2n} C_n = \left( \frac{3n}{2} - 1 \right) { }^{2n} C_n = \left( \frac{3n-2}{2} \right) { }^{2n} C_n$.

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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