Binomial Theorem 2 Question 8
11. Let $X=\left({ }^{10} C _1\right)^{2}+2\left({ }^{10} C _2\right)^{2}+3\left({ }^{10} C _3\right)^{2}+\ldots+10\left({ }^{10} C _{10}\right)^{2}$, where ${ }^{10} C _r, \quad r \in{1,2, \ldots, 10}$ denote binomial coefficients. Then, the value of $\frac{1}{1430} X$ is
(20̈18 Adv.)
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Answer:
Correct Answer: 11. (646)
Solution:
- We have,
$$ \begin{aligned} X & =\left({ }^{10} C _1\right)^{2}+2\left({ }^{10} C _2\right)^{2}+3\left({ }^{10} C _3\right)^{2}+\ldots+10\left({ }^{10} C _{10}\right)^{2} \\ \Rightarrow \quad X & =\sum _{r=1}^{10} r\left({ }^{10} C _r\right)^{2} \Rightarrow X=\sum _{r=1}^{10} r{ }^{10} C _r{ }^{10} C _r \\ \Rightarrow \quad X & =\sum _{r=1}^{10} r \times \frac{10}{r}{ }^{9} C _{r-1}{ }^{10} C _r \quad \because{ }^{n} C _r=\frac{n}{r}{ }^{n-1} C _{r-1} \end{aligned} $$
$$ \begin{aligned} & \Rightarrow X=10 \sum _{r=1}^{10}{ }^{9} C _{r-1}{ }^{10} C _r \\ & \Rightarrow X=10 \sum _{r=1}^{10}{ }^{9} C _{r-1}{ }^{10} C _{10-r} \quad\left[\because{ }^{n} C _r={ }^{n} C _{n-r}\right] \\ & \Rightarrow X=10 \times{ }^{19} C _9 \quad\left[\because{ }^{n-1} C _{r-1}{ }^{n} C _{n-r}={ }^{2 n-1} C _{n-1}\right] \\ & \text { Now, } \frac{1}{1430} X=\frac{10 \times{ }^{19} C _9}{1430}=\frac{{ }^{19} C _9}{143}=\frac{{ }^{19} C _9}{11 \times 13} \\ & =\frac{19 \times 17 \times 16}{8}=19 \times 34=646 \end{aligned} $$
