Question

# Find the sum of the products, two at a time, of the coefficients in the expansion of $$(1+x)^n$$ when $$n$$ is a positive integer.

Solution

## $$(1+x)^{n}=C_{0}+C_{1}x+C_{2}x^{2}.....C_{n}x^{n}$$ .............(1)squaring both sides and put x = 1$$(2)^{2n}=[C_{0}+C_{1}+C_{2}.....C_{n}]^{2}$$$$(2)^{2n}= [C_{0}^{2}+C_{1}^{2}+C_{3}^{2}+.......C_{n}^{2}] + 2.[C_{1}C_{2}+C_{2}C_{3}.....]$$let us assume sum taken two at a time be S$$S=\dfrac{(2)^{2n}- [C_{0}^{2}+C_{1}^{2}+C_{3}^{2}+.......C_{n}^{2}]}{2}$$replace    $$x$$   by $$\dfrac{x}{2}$$  from eqn (1)and multiply with eqn (1)$$(1+x)^{n}.(1+\dfrac{1}{x})^{n}=(1+x)^{2n}.\dfrac{1}{x^{n}}=[C_{0}+C_{1}x+C_{2}x^{2}.....C_{n}x^{n}].[C_{0}+C_{1}\dfrac{1}{x}+C_{2}\dfrac{1}{x^{2}}+.....C_{n}\dfrac{n}{x^{n}}]$$contant term at $$RHS = C_{0}^{2}+C_{1}^{2}+C_{3}^{2}+.......C_{n}^{2}$$contant term at $$LHS$$ = coefficient of  $$x^{n}$$ = constant term at RHS$$= C_{n}^{2n}=C_{0}^{2}+C_{1}^{2}+C_{3}^{2}+.......C_{n}^{2}$$ using above value $$S=\dfrac{(2)^{2n}- C_{n}^{2n}}{2}$$$$=2^{2n-1}- \dfrac{(2n)!}{2.n!.n!}$$Mathematics

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