SATHEE: Binomial Theorem 2 Question 16

Binomial Theorem 2 Question 16

19.

Prove that $C_{1}^{2} - 2 \cdot C_{2}^{2} + 3 \cdot C_{3}^{2} - \dots - 2 n \cdot C_{2 n}^{2} = (- 1)^{n} n \cdot C_{n}$

(1979, 4M)

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Solution:

We know that, $(1 + x)^{2 n} = C_{0} + C_{1} x + C_{2} x^{2} + \dots + C_{2 n} x^{2 n}$

On differentiating both sides w.r.t. $x$ , we get $2 n (1 + x)^{2 n - 1} = C_{1} + 2 \cdot C_{2} x + 3 \cdot C_{3} x^{2}$

$+ \dots + 2 n C_{2 n} x^{2 n - 1}$ …….(i)

and $1 - {\frac{1}{x}}^{2 n} = C_{0} - C_{1} \cdot \frac{1}{x} + C_{2} \cdot \frac{1}{x^{2}} - C_{3} \cdot \frac{1}{x^{3}}$

$+ \dots + C_{2 n} \cdot \frac{1}{x^{2 n}}$ …….(ii)

On multiplying Eqs. (i) and (ii), we get

$2 n (1 + x)^{2 n - 1} (1 - \frac{1}{x})^{2 n}$

$= [C_{1} + 2 \cdot C_{2} x + 3 \cdot C_{3} x^{2} + \dots + 2 n \cdot C_{2 n} x^{2 n - 1}]$

$\times [C_{0} - C_{1} \frac{1}{x} + C_{2} \frac{1}{x^{2}} - \dots . . + C_{2 n} \frac{1}{x^{2 n}}]$

Coefficient of $(\frac{1}{x})$ on the LHS

$\begin{aligned} = Coefficient of \frac{1}{x} in 2 n (\frac{1}{x^{2 n}}) (1 + x)^{2 n - 1} (x - 1)^{2 n} \\ = Coefficient of x^{2 n - 1} in 2 n {(1 - x^{2})}^{2 n - 1} (1 - x) \\ = 2 n (- 1)^{n - 1} \cdot (2 n - 1) C_{n - 1} (- 1) \\ = (- 1)^{n} (2 n) \frac{(2 n - 1)!}{(n - 1)! n!} = (- 1)^{n} n \frac{(2 n)!}{(n!)^{2}} \cdot n \\ = - (- 1)^{n} n \cdot C_{n} \dots \dots . (i i i) \end{aligned}$

Again, the coefficient of $(\frac{1}{x})$ on the RHS

$= - (C_{1}^{2} - 2 \cdot C_{2}^{2} + 3 \cdot C_{3}^{2} - \dots - 2 n C_{2 n}^{2})$ …….(iv)

From Eqs. (iii) and (iv),

$C_{1}^{2} - 2 \cdot C_{2}^{2} + 3 \cdot C_{3}^{2} - \dots - 2 n \cdot C_{2 n}^{2} = (- 1)^{n} n \cdot C_{n}$

Binomial Theorem 2 Question 16

19.

Solution:

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Binomial Theorem 2 Question 16

19.

Solution:

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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