If $C$ 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}

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Solution

The correct option is D $(\frac{3 n - 2}{2}) {^{2 n} C}_{n}$
${(1 + x)}^{n} =^{n} C_{0} +^{n} C_{1} x +^{n} C_{2} x^{2} + . . . +^{n} C_{n} x^{n}$
${(1 + \frac{1}{x})}^{n} =^{n} C_{0} + \frac{^{n} C_{1}}{x} + \frac{^{n} C_{2}}{x^{2}} + . . . + \frac{^{n} C_{n}}{x^{n}}$
${(1 + x)}^{n} {(1 + \frac{1}{x})}^{n} = (^{n} C_{0} +^{n} C_{1} x +^{n} C_{2} x^{2} + . . . +^{n} C_{n} x^{n}) (^{n} C_{0} + \frac{^{n} C_{1}}{x} + \frac{^{n} C_{2}}{x^{2}} + . . . + \frac{^{n} C_{n}}{x^{n}})$
In RHS ${^{n} C_{0}}^{2} + {^{n} C_{1}}^{2} + . . . + {^{n} C_{n}}^{2}$ comes as constant term.
So, ${^{n} C_{0}}^{2} + {^{n} C_{1}}^{2} + . . . + {^{n} C_{n}}^{2}$ is constant term in $(1 + x) {(1 + \frac{1}{x})}^{n}$
$= \frac{{(1 + x)}^{2 n}}{x^{n}}$
The constant term in $\frac{{(1 + x)}^{2 n}}{x^{n}}$ is $^{2 n} C_{n}$ as $\frac{x^{n}}{x^{n}}$ gets cancelled.
So, ${^{n} C_{0}}^{2} + {^{n} C_{1}}^{2} + . . . + {^{n} C_{n}}^{2} =^{2 n} C_{n}$
${(1 + x)}^{n} =^{n} C_{0} +^{n} C_{1} x +^{n} C_{2} x^{2} + . . . +^{n} C_{n} x^{n}$
by differentiating
$n {(1 + x)}^{n - 1} = {0.}^{n} C_{0} + {1.}^{n} C_{1} + {2.}^{n} C_{2} x + . . . + n .^{n} C_{n} x^{n - 1}$
${(1 + \frac{1}{x})}^{n} =^{n} C_{0} + \frac{^{n} C_{1}}{x} + \frac{^{n} C_{2}}{x^{2}} + . . . + \frac{^{n} C_{n}}{x^{n}}$
By multiplying
$n {(1 + x)}^{n - 1} \times {(1 + \frac{1}{x})}^{n} = (.^{n} C_{0} + {1.}^{n} C_{1} + {2.}^{n} C_{2} x + . . . + n^{n} C_{n} x^{n - 1}) \times (^{n} C_{0} + \frac{^{n} C_{1}}{x} + \frac{^{n} C_{2}}{x^{2}} + . . . + \frac{^{n} C_{n}}{x^{n}})$
So, $0. {^{n} C_{0}}^{2} + 1. {^{n} C_{1}}^{2} + 2. {^{n} C_{2}}^{2} + . . . + n . {^{n} C_{n}}^{2}$ is coefficient of $\frac{1}{n}$ in RHS
So, it is $\frac{1}{n}$ term in $n {(1 + x)}^{n - 1} \times \frac{{(1 + x)}^{n}}{x^{n}}$
i.e. $n .^{2 n - 1} C_{n - 1} = \frac{n}{2} .^{2 n} C_{n}$
So, $\sum {(3 n - 1) C_{n}}^{2} = 3 \sum {n C_{n}}^{2} - \sum {C_{n}}^{2}$
$= \frac{3 n}{2} .^{2 n} C_{n} -^{2 n} C_{n} = \frac{3 n - 2}{2} \times^{2 n} C_{n}$

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C_{r}

{^{n} C}_{r}

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

(1 + x)^{n} = n \sum r = 0^{n} C_{r} x^{n},

\frac{C_{0}}{1 \cdot 2} 2^{2} + \frac{C_{1}}{2 \cdot 3} 2^{3} + \frac{C_{2}}{3 \cdot 4} 2^{4} + \dots + \frac{C_{n}}{(n + 1) (n + 2)} 2^{n + 2}

lim n \to \infty \frac{5^{n + 1} + 3^{n} - 2^{2 n}}{5^{n} + 2^{n} + 3^{n + 2}}, n ϵ N .

L i m i t n \to \infty \frac{5^{n + 1} + 3^{n} - 2^{2 n}}{5^{n} + 2^{n} + 3^{2 n + 3}} =

C_{0} + \frac{3}{2} . C_{1} + C_{2} + \frac{27}{4} . C_{3} + + \frac{3^{n}}{n + 1} . C_{n} = \frac{4^{n + 1} + 1}{3 (n + 1)}

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