Download the app

Questions  

In the expansion of x31x2n,nN, if the sum of the coefficients of x5 and x10is 0, then n is

a
25
b
20
c
15
d
None of these

detailed solution

Correct option is C

x3−1r2nGeneral term Tr+1=n!r!(n−r)!(−1)n−rx5n−2nIf 5r−2n=5, then 5r=2n+5 or r=2n5+1If 5r−2n=10, then 5r=2n+10 or r=2n5+2x5 and x10 terms occurs if n=5kGiven that sum of x5 and x10 is zero.⇒5k!(2k+1)!(3k−1)!−5k!(2k+2)!(3k−2)!=0or  13k−1−12k+2=0or   k=3⇒n=15

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