Let \[ X=\left({ }^{10} C_{1}\right)^{2}+2\left({ }^{10} C_{2}\right)^{2}+3\left({ }^{10} C_{3}\right)^{2}+\cdots+10\left({ }^{10} C_{10}\right)^{2}, \] where ${ }^{10} C_{r}, r \in\{1,2, \cdots, 10\}$ denote binomial coefficients. Then, the value of $\frac{1}{1430} X$ is ____.

JEE Advanced · 2018 · 20 May 2018 · Paper 2 · Binomial Theorem

Let

X = (^{10} C_{1})^{2} + 2 (^{10} C_{2})^{2} + 3 (^{10} C_{3})^{2} + \dots + 10 (^{10} C_{10})^{2},

where

^{10} C_{r}, r \in {1, 2, \dots, 10}

denote binomial coefficients. Then, the value of

\frac{1}{1430} X

is ____.

Answer: 646

Solution

Using the identity:

r = 1 \sum n r (^{n} C_{r})^{2} = n \cdot^{2 n - 1} C_{n - 1}

For

n = 10

X = 10 \cdot^{19} C_{9}

\frac{X}{1430} = \frac{10 \cdot ^{19} C _{9}}{1430}

Calculating

^{19} C_{9} = \frac{19 !}{9 ! \cdot 10 !} = 92378

\frac{10 \cdot 92378}{1430} = \frac{923780}{1430} = 646

JEE Advanced · 2018 · 20 May 2018 · Paper 2 · Numerical answer

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