Q.

If the sum of the coefficients of x7 and x14 in the expansion of (1x3–x4)n, x≠0, is zero, then the value of n is ________.         [2026]


Ans.

(21)

We have, (1x3–x4)n

General term is given by

Tr+1=Crn(1x3)n–r(–x4)r

              =Crnx–3n+3r+4r(–1)r=Crnx–3n+7r(–1)r

∴  –3n+7r=7 ⇒7r=7+3n

⇒ r7+3n7 ⇒r=1+3n7

∴  –3n+7r=14 ⇒ 7r=14+3n

⇒ r=2+3n7

Since, Coefficient of x7 + Coefficient of x14=0

∴  C(1+3n7)n(–1)1+3n7+C(2+3n7)n(–1)2+3n7=0

⇒ (–1)1+3n7[C(1+3n7)n–C(2+3n7)n]=0

⇒C(1+3n7)n=C(2+3n7)n

⇒ 1+3n7+2+3n7=n ⇒ 7n–6n7=3

n=21