Question

求(1+2x-3x²)⁶的展開式中含x⁵的項.

Answer 1

解析：∵(1+2x-3x²)⁶=(3x²-2x-1)⁶=［(x-1)(3x+1)］⁶=(x-1)⁶(3x+1)⁶,

而(1+3x)⁶的通項為T_k+1=3^k··x^k,

(x-1)⁶=(1-x)⁶的通項為T_r+1=(-1)^r··x^r，

∴(x-1)⁶(3x+1)⁶的通項為(-1)^r··3^k··x^r+k.

令r+k=5且k∈{0,1,2,3,4,5,6},r∈{0,1,2,3,4,5,6},

∴

故所求的項為

［·3⁵·+(-1)¹··3⁴·+(-1)²··3³·+(-1)³··3²·+(-1)⁴··3¹·+(-1)⁵··3⁰·］x⁵=-168x⁵.