SATHEE: Binomial Theorem 2 Question 17

Binomial Theorem 2 Question 17

20.

Prove that ${(^{2 n} C_{0})}^{2} - {(^{2 n} C_{1})}^{2} + {(^{2 n} C_{2})}^{2} - \dots + {(^{2 n} C_{2 n})}^{2}$ $= (- 1)^{n} \cdot^{2 n} C_{n}$ .

$(1978, 4$ M)

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Solution:

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

$\begin{aligned} = & [^{2 n} C_{0} + (^{2 n} C_{1}) x + (^{2 n} C_{2}) x^{2} + \dots + (^{2 n} C_{2 n}) x^{2 n}] \\ \times [^{2 n} C_{0} - (^{2 n} C_{1}) \frac{1}{x} + (^{2 n} C_{2}) \frac{1}{x^{2}} + \dots + (^{n} C_{2 n}) \frac{1}{x^{2 n}}] \end{aligned}$

Independent terms of $x$ on RHS

$\begin{aligned} = {(^{2 n} C_{0})}^{2} - {(^{2 n} C_{1})}^{2} + {(^{2 n} C_{2})}^{2} - \dots + {(^{2 n} C_{2 n})}^{2} \\ LHS & = (1 + x)^{2 n} (\frac{x - 1}{x})^{2 n} = \frac{1}{x^{2 n}} {(1 - x^{2})}^{2 n} \end{aligned}$

Independent term of $x$ on the LHS $= (- 1)^{n} \cdot^{2 n} C_{n}$ .

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Binomial Theorem 2 Question 17

20.

Solution:

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