If the term free from x in the expansion of (√−kx2)10 is 405, find the value of k.
Let (r+1)th term, in the expansion of (√−kx2)10, be equal to Tr+1=10Cr(√)10−r(−kx2)r
If Tr+1 is independent of x, then 5−5r2=0
Putting r =2 in(i), we obtain
But it is given that the value of the term free from x is 405
Hence, the value of k is ±3.