3r th term in the expansion of (1+x)2n =2nC3r−1x3r−1
and (r+2) th term in the expansion of (1+x)2n =2nCr+1xr+1
Given that the binomial coefficients of (3r) th
and (r+2) th terms are equal. ∴2nC3r−1=2nCr+1 ⇒3r−1=r+1
or 2n=(3r−1)+(r+1) ⇒2r=2 or 2n=4r ⇒r=1 or n=2r
But r>1 ∴n=2r