If T2T3 in the expansion of (a+b)n and T3T4 in the expansion of (a+b)n+3 are equal, then n =
5
In (a+b)n in T2T3 = nC1an−1bnC2an−2b2 = 2a(n−1)b and (a+b)n+3 T3T4 = n+3C2an+1b2n+3C3anb3 = 3a(n+1)b
If these are equal then 2a(n−1)b = 3a(n+1)b so n = 5