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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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