Binomial Theorem 2 Question 15
18.
If $(1+x)^{n}=C_{0}+C_{1} x+C_{2} x^{2}+\ldots+C_{n} x^{n}$, then show that the sum of the products of the $C_{i}^{\prime}$ s taken two at a time represented by $\Sigma \Sigma C_{i} C_{j}$ 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}-\left({ }^{2 n} C_{n}\right)=2^{2 n}-{ }^{2 n} C_{n} \\ \therefore \quad \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} $
