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  4. Prove that (2nC0)2- (2nC1)2+(2nC2)2-....+(2nC2n)2 =(-1)n. 2nCn

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Q. Prove that $(^{2n}C_0)^2- (^{2n}C_1)^2+(^{2n}C_2)^2-....+(^{2n}C_{2n})^2 =(-1)^n. \,{}^{2n}C_n$

IIT JEEIIT JEE 1978Binomial Theorem

Solution:

$(1+x)^2n \bigg(1-\frac{1}{x}\bigg)^{2n}$
$=[(^{2n}C_0)^2- (^{2n}C_1)^2+(^{2n}C_2)^2+...+(^{2n}C_{2n})^x{^{^{2n}}}]$
$\times \bigg[ \,{}^{2n}C_0-(^{2n}C_1)\frac{1}{x}+(^{2n}C_2)\frac{1}{x^2}+...+(^{2n}C_{2n}) \frac{1}{x^{2n}} \bigg]$
Independent terms of $x$ on RHS
$=(^{2n}C_0)^2 - (^{2n}C_1)^2 + (^{2n}C_2)^2 -,,,+(^{2n}C_{2n})^2$
LHS =$(1+x)^{2n} \bigg(\frac{x-1}{x}\bigg)^{2n} = \frac{1}{x^{2n}}(1-x^2)^{2n}$
Independent term of $x$ on the LHS =$(-1)^n. \,{}^{2n}C_n.$