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,
$2 \sum \sum _{0 \leq i<j \leq n} C _i C _j=\sum _{i=0}^{n} \sum _{j=0}^{n} C _i C _j-\sum _{i=0}^{n} \sum _{j=0}^{n} C _i C _j$
$$ \begin{aligned} & =\sum _{i=0}^{n} C _i \sum _{j=0}^{n} C _j-\sum _{i=0}^{n} C _i^{2} \\ & =2^{n} 2^{n}-\left({ }^{2 n} C _n\right)=2^{2 n}-{ }^{2 n} C _n \\ \therefore \quad \sum _{0 \leq i<j \leq n} C _i C _j & =\frac{2^{2 n}-{ }^{2 n} C _n}{2}=2^{2 n-1}-\frac{(2 n) !}{2(n !)^{2}} \end{aligned} $$
