Download the app

Questions  

In the expansion of x31x215, the constant term, is 

a
15C9
b
0
c
-15C9
d
1

detailed solution

Correct option is C

Let (r+1)th  term be the constant term in the  expansion of x3−1x215∴ Tr+1=15Crx315−r−1x2r  is independent of x⇒Tr+1=15Crx45−5r(−1)r is independent of x ⇒45−5r=0⇒r=9 Thus, tenth term is independent of x and is given byT10=15C9(−1)9=−15C9

Talk to our academic expert!

+91

Are you a Sri Chaitanya student?


Similar Questions

lf the term independent of x in the expansion of 32x213x9is k, then 18k is equal to


phone icon
whats app icon