QJEEAdv18Ma1a5d: JEE Advanced 2018 Mathematics - Class 11 JEEAdv18Ma1a5d Solution: Let , where , r {1, 2, ..., 10} denote binomial coefficients.…

Question

Verified Solution • 2025–26 Session moderate

Let

X = {({}^{10}{C_1})^2} + 2{({}^{10}{C_2})^2} + 3{({}^{10}{C_3})^2} + ... + 10{({}^{10}{C_{10}})^2}

,

where

{}^{10}{C_r}

, r

\in

{1, 2, ..., 10} denote binomial coefficients. Then, the value of

{1 \over {1430}}X

is ..........

Accepted Answer

We have,

X = {({}^{10}{C_1})^2} + 2{({}^{10}{C_2})^2} + 3{({}^{10}{C_3})^2} + ... + 10{({}^{10}{C_{10}})^2}

\Rightarrow X = \sum\limits_{r = 1}^{10} r {({}^{10}{C_r})^2}

\Rightarrow X = \sum\limits_{r = 1}^{10} r\times {}^{10}{C_r}\times{}^{10}{C_r}

\Rightarrow X = \sum\limits_{r = 1}^{10} r \times {}^{10}{C_r}\times {{10} \over r}\times{}^9{C_{r - 1}}

\left[ \,\because{{}^{n}{C_r} = {n \over r}{}^{n - 1}{C_{r - 1}}} \right]

\Rightarrow X = 10\sum\limits_{r = 1}^{10} {{}^9{C_{r - 1}}} {}^{10}{C_r}

\Rightarrow X = 10\sum\limits_{r = 1}^{10} {{}^9{C_{r - 1}}} {}^{10}{C_{10 - r}}

[

\because

{}^n{C_r} = {}^n{C_{n - r}}

]

\Rightarrow X = 10 \times {}^{19}{C_9}

[

\because

{}^{n - 1}{C_{r - 1}}{}^n{C_{n - r}} = {}^{2n - 1}{C_{n - 1}}

]

Now,

{1 \over {1430}}X = {{10 \times {}^{19}{C_9}} \over {1430}} = {{{}^{19}{C_9}} \over {143}} = {{{}^{19}{C_9}} \over {11 \times 13}}

= {{19 \times 17 \times 16} \over 8} = 19 \times 34 = 646

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