Download the app

Questions  

If in the expansion of  x31x2nthe 

sum of the coefficients of x5 and  x10 is  0

 then the coefficient of x20is

a
20C6
b
−20C6
c
15C5
d
−15C5

detailed solution

Correct option is D

(r + 1)th term in the expansion ofx3−1x2n is Tr+1=nCrx3n−r−1x2r=nCrx3n−5r(−1)rFor coefficient of r10, set  3n−5r=10⇒ r=35n−1=r1  (say) Note that, r1=r2+1We are given nCr1(−1)r1+nCr2(−1)r2=0 nCr2=nCr2+1∵r1=r2+1⇒r2+r2+1=n⇒r2=12(n−1)⇒ 35n−2=12n−12⇒110n=32⇒n=15For coefficient of x20 set  3n−5r=205r=45−20=25 or  r=5 Thus, coefficient of x20 is  −15C5

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