If  n−1Cr=k2−3nCr+1,, then k belongs to

We have,  n−1Cr=k2−3nCr+1⇒(n−1)!(n−1−r)!r!=k2−3n!(n−r−1)!(r+1)!⇒1=k2−3n(r+1)⇒r+1n=k2−3                    [∵0≤r≤n−1⇒r+1≤n]⇒k2−3≤1⇒k2−4≤0⇒−2≤k≤2      ... (i)Again  n−1Cr=k2−3nCr+1⇒ k2−3= n−1Cr nCr+1>0⇒ k2−3>0⇒k<−3or  k>3              ... (ii)From (i) and (ii), we have k∈(3,2].