✔ 最佳答案
With the exponential function expressed in Maclaurin's series, it is:
ex = 1 + x + x2/2! + x3/3! + ... + xn/n! + ...
So the coeff. in (2x + 1)ex is:
(2x + 1)ex = (2x + 1)(1 + x + x2/2! + x3/3! + ... + xn-1/(n - 1)! + xn/n! + ...)
Coeff. of xn:
2/(n - 1)! + 1/n!
= 2n/n! + 1/n!
= (2n + 1)/n!
2011-02-10 15:51:41 補充:
1) 2x 乘以 e^x 入面的 x^(n-1), 得出 2/(n - 1)!
2) 1 乘以 e^x 入面的 x^n, 得出 1/n!
將兩者相加即可
2011-02-10 17:16:04 補充:
2x 乘入去 x^(n-1)/(n-1)! = 2x^n/(n-1)!
1 乘入去 x^n/n! = x^n/(n-1)!
兩個 terms 皆為 x^n 的, 所以加埋就得出 coeff.
同埋2x駛唔駛再乘入去x^(n)/n! ?
不用, 因為會得出 2x^(n + 1)/n!, 不是 x^n 的 term