Question

请教一道高数题？

Answer 1

cos2x 展开的通项：{(2x)^(2n) (-1)^n/(2n)!}，[0,oo)
-1/(1+x) 展开的通项: -(-1)^n x^n, [0,oo)
求导：1/(1+x)^2 展开的通项: (-1)^n nx^(n-1), [1,oo) = (-1)^n (n+1)x^n, [0,oo)
cos2x - 1/(1+x)^2 = {(2x)^(2n) (-1)^n/(2n)!}，[0,oo) - {(-1)^n (n+1)x^n}, [0,oo)
n = odd, an = -(-1)^n (n+1);
n = even, an = 2^n (-1)^(n/2)/n! - (n+1), 注意: (-1)^(n/2)
取前三项验证：cos2x - 1/(1+x)^2 = (1 - (2x)^/2! + (2x)^4/4! +...) - (1 - 2x + 3x^2)
a0 = 0 = 2^0 (-1)^(0/2)/0! - (0+1) 满足
a1 = 2 = -(-1)^1 (1+1) 满足
a2 = -2^2/2! - 3 = 2^2 (-1)^(2/2)/2! - (2+1) 满足

Answer 2

Answer 3

cos2x要写成x^n形式，而不能写成原来的x^2n的形式，不然形式上不统一。