如果|x|<=1, 试求(1 + x + x^2 + x^3)^-1 展开式中的x^13及x^16的系数?
回答 (1)
2016-07-03 5:23 pm
✔ 最佳答案
( 1 + x + x^2 + x^3 )^(-1)= [ ( 1 - x^4 ) / ( 1 - x ) ]^(-1)
= ( 1 - x ) / ( 1 - x^4 )
= ( 1 - x ) * [ 1 / ( 1 - x^4 ) ]
= ( 1 - x )( 1 + x^4 + x^8 + x^12 + x^16 + ..... ) , 因為 |x| ≦ 1
x^13 項 = - x * x^12 = - x^13
x^13 係數 = - 1 ..... Ans
x^16 項 = 1 * x^16 = x^16
x^16 係數 = 1 ..... Ans
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