SATHEE: Binomial Theorem 2 Question 15

Binomial Theorem 2 Question 15

18.

If $(1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + \dots + C_{n} x^{n}$ , then show that the sum of the products of the $C_{i}^{'}$ s taken two at a time represented by $Σ Σ C_{i} C_{j}$ is equal to

$0 \leq i < j \leq n 2^{2 n - 1} - \frac{(2 n!)}{2 (n!)^{2}}$

(1983, 3M)

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Solution:

We know that,

$2 \sum \sum_{0 \leq i < j \leq n} C_{i} C_{j} = \sum_{i = 0}^{n} \sum_{j = 0}^{n} C_{i} C_{j} - \sum_{i = 0}^{n} \sum_{j = 0}^{n} C_{i} C_{j}$

$\begin{aligned} = \sum_{i = 0}^{n} C_{i} \sum_{j = 0}^{n} C_{j} - \sum_{i = 0}^{n} C_{i}^{2} \\ = 2^{n} 2^{n} - (^{2 n} C_{n}) = 2^{2 n} -^{2 n} C_{n} \\ ∴ \sum_{0 \leq i < j \leq n} C_{i} C_{j} & = \frac{2^{2 n} -^{2 n} C_{n}}{2} = 2^{2 n - 1} - \frac{(2 n)!}{2 (n!)^{2}} \end{aligned}$

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Binomial Theorem 2 Question 15

18.

Solution:

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