SATHEE: Binomial Theorem 2 Question 8

Binomial Theorem 2 Question 8

11.

Let $X = {(^{10} C_{1})}^{2} + 2 {(^{10} C_{2})}^{2} + 3 {(^{10} C_{3})}^{2} + \dots + 10 {(^{10} C_{10})}^{2}$ , where $^{10} C_{r}, r \in 1, 2, \dots, 10$ 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,

$\begin{aligned} X & = {(^{10} C_{1})}^{2} + 2 {(^{10} C_{2})}^{2} + 3 {(^{10} C_{3})}^{2} + \dots + 10 {(^{10} C_{10})}^{2} \\ \Rightarrow X & = \sum_{r = 1}^{10} r {(^{10} C_{r})}^{2} \Rightarrow X = \sum_{r = 1}^{10} r^{10} C_{r}^{10} C_{r} \\ \Rightarrow X & = \sum_{r = 1}^{10} r \times \frac{10}{r}^{9} C_{r - 1}^{10} C_{r} [∵^{n} C_{r} = \frac{n}{r}^{n - 1} C_{r - 1}] \end{aligned}$

$\begin{aligned} \Rightarrow X = 10 \sum_{r = 1}^{10}^{9} C_{r - 1}^{10} C_{r} \\ \Rightarrow X = 10 \sum_{r = 1}^{10}^{9} C_{r - 1}^{10} C_{10 - r} [∵^{n} C_{r} =^{n} C_{n - r}] \\ \Rightarrow X = 10 \times^{19} C_{9} [∵^{n - 1} C_{r - 1}^{n} C_{n - r} =^{2 n - 1} C_{n - 1}] \\ Now, \frac{1}{1430} X = \frac{10 \times^{19} C_{9}}{1430} = \frac{^{19} C_{9}}{143} = \frac{^{19} C_{9}}{11 \times 13} \\ = \frac{19 \times 17 \times 16}{8} = 19 \times 34 = 646 \end{aligned}$

Binomial Theorem 2 Question 8

11.

Answer:

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Binomial Theorem 2 Question 8

11.

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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