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} $
