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

Question

Prove that (2nC0)2- (2nC1)2+(2nC2)2-....+(2nC2n)2 =(-1)n. 2nCn (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

(1+x)2n (1-(1/x))2n =[(2nC0)2- (2nC1)2+(2nC2)2+...+(2nC2n)x2n] × [ 2nC0-(2nC1)(1/x)+(2nC2)(1/x2)+...+(2nC2n) (1/x2n) ] Independent terms of x on RHS =(2nC0)2 - (2nC1)2 + (2nC2)2 -,,,+(2nC2n)2 LHS =(1+x)2n ((x-1/x))2n = (1/x2n)(1-x2)2n Independent term of x on the LHS =(-1)n. 2nCn.

Q. Prove that (2nC0​)2−(2nC1​)2+(2nC2​)2−....+(2nC2n​)2=(−1)n.2nCn​

Q. Prove that $(^{2 n} C_{0})^{2} - (^{2 n} C_{1})^{2} + (^{2 n} C_{2})^{2} - .... + (^{2 n} C_{2 n})^{2} = (- 1)^{n} .^{2 n} C_{n}$