In the expansion of (x³- 1 x² )^ n where n is a positive integer, the sum of the coefficients of x ⁵ and x ¹⁰ is 0 . What is the sum of the coefficients of the two middle terms?

Step-by-step solution

(x³- 1 x² )^ n General term, T_ r+1 = n r (x³ )^ n-r (- 1 x² )^ r = n r x^ (3 n-3 r) (-1)^ r x^ -2 r = n r (-1)^ r x^ (3 n-5 r) ...(i) For the coefficient x ⁵ Put 3 n-5 r=55 r=3 n-5 r = 3 n 5 -1 Coefficient of x⁵= ^ n C ( 3 n 5 -1 )^ (-1) ( 3 n 5 -1 ) For the coefficient of x ¹⁰ Put 3 n-5 r=105 r=3 n-10 r = 3 n 5 -2 Coefficient of x¹⁰= n ( 3 n 5 -2 )^ (-1) ( 3 n 5 -2 ) The sum of the coefficient of x⁵ and x¹⁰=0 ^ n C _ ( 3 n 5 -1 )^ (-1) ( 3 n 5 -1 ) + ^ n C _ ( 3 n 5 -2 ) (-1)^ ( 3 n 5 -2 ) =0 .(-1)^ 3 n 5 [ n ( 3