1.

Prove that `(""^(2n)C_(0))^2-(""^(2n)C_(1))^2+(""^(2n)C_(2))^2-.....+(-1)^n(""^(2n)C_(2n))^2=(-1)^n.""^(2n)C_(n)`

Answer» `(""^(2n)C_(0))^3-(""^(2n)C_(1))^3-(""^(2n)C_(2))^3-.....+(-1)^n""^(2n)C_(2n)x^(2n)....(1)`
and `(x+1)^(2n)=""^(2n)C_(0)x^(2n)+""^(2n)C_(1)x^(2n-1)+""^(2n)C_(2)x^(2n-2)+....+""^(2n)C_(2n)...(ii)` Multiplying (i) and (ii) ,we get
`(x^2-1)^(2n)=(""^(2n)C_(0)-""^(2n)C_(1)x+....+(-1)^n""^(2n)C_(2n))xx(""^(2n)C_(0)x^(2n)+""^(2n)C_(1)x^(2n-1)+.....+""^(2n)C_(2n))`
Now ,cofficient of f`x^(2n)` in RHS
`=(""^(2n)C_(0))^2-(""^(2n)C_(1))^2+(""^(2n)C_(2))^2-......+(-1)^n(""^(2n)C_(2n)^2`
`therefore "General term in L.H.S "., T_(r+1)=""^(2n)C_(r)(x^2)C_(r)(x^2)^(2n-r)(-1)^r`


Discussion

No Comment Found

Related InterviewSolutions