JEE Advance - Mathematics (2018 - Paper 2 Offline - No. 14)
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 ..........
where $${}^{10}{C_r}$$, r $$ \in $${1, 2, ..., 10} denote binomial coefficients. Then, the value of $${1 \over {1430}}X$$ is ..........
Answer
646
Explanation
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, $${1 \over {1430}}X = {{10 \times {}^{19}{C_9}} \over {1430}} = {{{}^{19}{C_9}} \over {143}} = {{{}^{19}{C_9}} \over {11 \times 13}}$$
$$ = {{19 \times 17 \times 16} \over 8} = 19 \times 34 = 646$$
$$X = {({}^{10}{C_1})^2} + 2{({}^{10}{C_2})^2} + 3{({}^{10}{C_3})^2} + ... + 10{({}^{10}{C_{10}})^2}$$
$$ \Rightarrow X = \sum\limits_{r = 1}^{10} r {({}^{10}{C_r})^2}$$
$$ \Rightarrow X = \sum\limits_{r = 1}^{10} r\times {}^{10}{C_r}\times{}^{10}{C_r}$$
$$ \Rightarrow X = \sum\limits_{r = 1}^{10} r \times {}^{10}{C_r}\times {{10} \over r}\times{}^9{C_{r - 1}}$$ $$\left[ \,\because{{}^{n}{C_r} = {n \over r}{}^{n - 1}{C_{r - 1}}} \right]$$
$$ \Rightarrow X = 10\sum\limits_{r = 1}^{10} {{}^9{C_{r - 1}}} {}^{10}{C_r}$$
$$ \Rightarrow X = 10\sum\limits_{r = 1}^{10} {{}^9{C_{r - 1}}} {}^{10}{C_{10 - r}}$$ [ $$ \because $$ $${}^n{C_r} = {}^n{C_{n - r}}$$]
$$ \Rightarrow X = 10 \times {}^{19}{C_9}$$
[ $$ \because $$ $${}^{n - 1}{C_{r - 1}}{}^n{C_{n - r}} = {}^{2n - 1}{C_{n - 1}}$$]
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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