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JEE Advanced 2018Paper 2
Exam
JEE Advanced
Year
2018
Paper
Paper 2
Subject
Mathematics
Chapter
Binomial Theorem

JEE Advanced 2018 · Paper 2 · Mathematics · Binomial Theorem

Let X = ( ^ 10 C _ 1 ) ^ 2 + 2 ( ^ 10 C _ 2 ) ^ 2 + 3 ( ^ 10 C _ 3...

Let X=(10C1)X = {\left({}^{10}{C}_{1}\right)}^{2} + 2{\left({}^{10}{C}_{2}\right)}^{2} + 3{\left({}^{10}{C}_{3}\right)}^{2} + \ldots + 10{\left({}^{10}{C}_{10}\right)}^{2} , where 10Cr,r∈{1,2,….,10}{}^{10}{C}_{r}, r \in \left\{1,2, \ldots . , 10\right\} denote binomial coefficients. Then, the value of 11430X\dfrac{1}{1430}X is ______ .

Correct answer

0

Step-by-step solution

X=∑r=0nr.(nCr)2;n=10X = \mathop{∑}_{r = 0}^{n}r .{\left(n{C}_{r}\right)}^{2}; n = 10
X=n.∑r=0nnCr.n−1Cr−1X = n . \mathop{∑}_{r = 0}^{n}n{C}_{r} . n - 1{C}_{r - 1}
X=n.∑r=1nnCn−r.n−1Cr−1X = n . \mathop{∑}_{r = 1}^{n}n{C}_{n - r} . n - 1{C}_{r - 1}
X=n.2n−1Cn−1;n=10X = n . 2n - 1{C}_{n - 1}; n = 10
X=10.19C9X = 10 . 19{C}_{9}
X1430=1143.19C9\dfrac{X}{1430} = \dfrac{1}{143} . 19{C}_{9}
=646= 646
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