函數運算 希望能幫助解答,萬分感激!~

(c)

h(n + 1)
= 2(n + 1)^2 - (n + 1)
= 2(n^2 + 2n + 1) - (n + 1)
= 2n^2 + 4n + 2 - n - 1
= 2n^2 + 3n + 1

(d)

h(n + 1) - h(n)
= (2n^2 + 3n + 1) - (2n^2 - n)
= 2n^2 + 3n + 1 - 2n^2 + n
= 4n + 1

(e)

h(x - 1)
= 2(x - 1)^2 - (x - 1)
= 2(x^2 - 2x + 1) - (x - 1)
= 2x^2 - 4x + 2 - x + 1
= 2x^2 - 5x + 3

建議先factorize h(x)。
h(x) = 2x²-x = x(2x-1)


c)
h(n+1)
= (n+1)[2(n+1)-1]
=(n+1)(2n+1)

d)
h(n+1) - h(n)
= (n+1)(2n+1) - n(2n-1)
= (2n²+3n+1) - (2n²-n)
= 4n+1

e)
h(x-1)
= (x-1)[2(x-1)-1]
= (x-1)(2x-3)

2013-01-15 14:21:47 補充：
其實part d我就咁做既:
h(n+1) - h(n)
= (n+1)(2n+1) - n(2n-1)
= n(2n+1) +(2n+1) -n(2n-1)
= n[(2n+1)-(2n-1)] + (2n+1)
= 4n+1

SORRY 我打"溜"左...h(x)=2x^2-x

你確定是"(x)"而不是"h(x)？