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]
(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