In the expansion of 1x2−x3n,n∈N, if the sum of the coefficients of x5 and x10 is 0, then n is

1x2−x3nGeneral term Tr+1=ncr1x2n-r-x3r=n!r!(n−r)!(−1)n−rx5r−2n for coefficient x5   5r - 2n = 5, then 5r = 2n + 5 or r=2n5+1…………(I)for coefficient  x10 5r - 2n = 10, then 5r = 2n + 10 or r=2n5+2……………………(II)x5 and x10 terms occurs if n = 5kin (I) put n = 5k  ⇒r = 2k+1in  (II) put n = 5k ⇒r = 2k+2Given that sum of x5 and x10 is zero.⇒ 5k!(2k+1)!(3k−1)!−5k!(2k+2)!(3k−2)!=0 or 13k−1−12k+2=0 or  k=3⇒n=15