If in the expansion of (a−2b)n,the sum of 4th and 5th term is zero, then the value of ab is

Here,T4=nC3(a)n−3(−2b)3and T5=nC4(a)n−4(−2b)4Given T4+T5=0⇒ nC3(a)n−3(−2b)3+nC4(a)n−4(−2b)4=0⇒ (a)n−4(−2b)3anC3+nC4(−2b)=0⇒ ab=2nC4 nC3=2⋅n(n−1)(n−2)(n−3)4⋅3⋅2⋅1×3⋅2⋅1n(n−1)(n−2)=n−32