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 .
Answer
21
Solution
Tr+1=nCr (x-3 )n-r (-x4 )r=(-1)r nCr x7r-3n
Let x7 come from Tr+1: 7r-3n=7; then x14 comes from Tr+2 (power rises by 7)
Sum of coefficients =(-1)r nCr+(-1)r+1 nCr+1=0
nCr=nCr+1 r+(r+1)=n n=2r+1
7r-3(2r+1)=7 r=10, n=21
Answer: 21