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Question

If the term free from x in the expansion of (xkx2)10 is 405, find the value of k.


Solution

Let (r+1)th term, in the expansion of (xkx2)10, be equal to Tr+1=10Cr(r)10r(kx2)r

=10Crx55r2(l)r....(i)

If Tr+1 is independent of x, then 55r2=0

r=2

Putting r =2 in(i), we obtain

T3=10C2(k)2=45k2

But it is given that the value of the term free from x is 405

45k2=405k2=9k=±3

Hence, the value of k is ±3.


Mathematics
RD Sharma
Standard XI

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