If $C_{r}$ denotes the binomial coefficient ${^{n} C}_{r}$ then $(- 1) C_{0}^{2} + 2 C_{1}^{2} + 5 C_{2}^{2} + . . . . . . (3 n - 1) C_{n}^{2} =$

(3 n - 2) {^{2 n} C}_{n}

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(\frac{3 n - 2}{2}) {^{2 n} C}_{n}

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(5 + 3 n) {^{2 n} C}_{n}

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(\frac{3 n - 5}{2}) {^{2 n} C}_{n + 1}

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Solution

The correct option is B $(\frac{3 n - 2}{2}) {^{2 n} C}_{n}$
The general term of above series is $T_{r} = (3 r - 1) C_{r}^{2}$

$(- 1) C_{0}^{2} + 2 C_{1}^{2} + 5 C_{2}^{2} + . . . + (3 n - 1) C_{n}^{2}$

$= \sum_{r = 0}^{r = n} (3 r - 1) C_{r}^{2}$

$= 3 \sum_{r = 0}^{r = n} r C_{r}^{2} - \sum_{r = 0}^{r = n} C_{r}^{2}$

$= 3 S_{1} - S_{2}$

The binomial expansion of $(1 + x)^{n}$ and $(1 + \frac{1}{x})^{n}$ is given by

$(1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + . . . . + C_{n} x^{n} = \sum_{r = 0}^{r = n} C_{r} x^{r}$ ....[1]
$(1 + \frac{1}{x})^{n} = C_{0} + C_{1} \frac{1}{x} + C_{2} \frac{1}{x^{2}} + . . . . + C_{n} \frac{1}{x^{n}} = \sum_{r = 0}^{r = n} C_{r} \frac{1}{x^{r}}$ ....[2]

To find the value of $S_{2}$ , we can multiply above two equations and then compare the coefficients of terms independent of x.
Multiplying equation [1] and [2], we get
$(1 + x)^{n} (1 + \frac{1}{x})^{n} = \sum_{r = 0}^{r = n} C_{r} x^{r} . \sum_{r = 0}^{r = n} C_{r} \frac{1}{x^{r}}$

$⟹ \frac{(1 + x)^{2 n}}{x^{n}} = \sum_{r = 0}^{r = n} C_{r} x^{r} . \sum_{r = 0}^{r = n} C_{r} \frac{1}{x^{r}}$
$⟹ \frac{(1 + x)^{2 n}}{x^{n}} = C_{0}^{2} + C_{1}^{2} + C_{2}^{2} + . . . + C_{n}^{2}$ + terms containing x
Therefore, comparing coefficients of $x^{0}$ in L.H.S and R.H.S, we get
Coefficients of $x^{0}$ in R.H.S = Coefficients of $x^{0}$ in L.H.S
$⟹ C_{0}^{2} + C_{1}^{2} + C_{2}^{2} + . . . + C_{n}^{2} =$ coefficient of $x^{0}$ in $\frac{(1 + x)^{2 n}}{x^{n}}$
$⟹ C_{0}^{2} + C_{1}^{2} + C_{2}^{2} + . . . + C_{n}^{2} =^{2 n} C_{n} = S_{2}$ ....[3]

$(1 + \frac{1}{x})^{n} = C_{0} + C_{1} \frac{1}{x} + C_{2} \frac{1}{x^{2}} + . . . . + C_{n} \frac{1}{x^{n}}$
Differentiating with respect to x, we get
$⟹ n (1 + \frac{1}{x})^{n - 1} (\frac{- 1}{x^{2}}) = 0 + C_{1} \frac{- 1}{x^{2}} + C_{2} \frac{- 2}{x^{3}} + . . . . + C_{n} \frac{- n}{x^{n + 1}}$
$⟹ n (1 + \frac{1}{x})^{n - 1} (\frac{1}{x}) = C_{1} \frac{1}{x} + C_{2} \frac{2}{x^{2}} + . . . . + C_{n} \frac{n}{x^{n}}$ ....[4]

To find the value of $S_{1}$ , we can multiply equations [1] and [4], and then compare the coefficients of terms independent of x.
Multiplying equation [1] and [4], we get
$n (1 + \frac{1}{x})^{n - 1} (\frac{1}{x}) (1 + x)^{n} = (C_{1} \frac{1}{x} + C_{2} \frac{2}{x^{2}} + . . . . + C_{n} \frac{n}{x^{n}}) (C_{0} + C_{1} x + C_{2} x^{2} + . . . . + C_{n} x^{n})$
$⟹ n \frac{(x + 1)^{2 n - 1}}{x^{n}} = (C_{1} \frac{1}{x} + C_{2} \frac{2}{x^{2}} + . . . . + C_{n} \frac{n}{x^{n}}) (C_{0} + C_{1} x + C_{2} x^{2} + . . . . + C_{n} x^{n})$
Therefore, comparing coefficients of $x^{0}$ in L.H.S and R.H.S, we get
Coefficients of $x^{0}$ in R.H.S = Coefficients of $x^{0}$ in L.H.S
$⟹ C_{1}^{2} + 2 C_{2}^{2} + 3 C_{3}^{2} + . . . + n C_{n}^{2} =$ coefficient of $x^{0}$ in $n \frac{(x + 1)^{2 n - 1}}{x^{n}}$
$⟹ C_{1}^{2} + 2 C_{2}^{2} + 3 C_{3}^{2} + . . . + n C_{n}^{2} = n (^{2 n - 1} C_{n}) = S_{1}$
Therefore,
$3 S_{1} - S_{2} = 3 n (^{2 n - 1} C_{n}) -^{2 n} C_{n}$
$= 3 n \frac{(^{2 n} C_{n})}{2} -^{2 n} C_{n}$
$= (\frac{3 n - 2}{2})^{2 n} C_{n}$

Hence, the answer is option (B)

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