F4 maths function

As f(x+1) - f(x) = 2x + 1, so f(x) is a quadratic function,
Let f(x) = ax^2 + bx + c, so f(x+1) = ax^2 + (2a+b)x + (a+b+c)
f(x+1) - f(x) = 2ax + (a+b) = 2x + 1,
So, a = 1, b = 0
ie. f(x) = x^2 + c

f(3) - f(1)
= (3^2 + c) - (1^2 + c)
= 9 - 1
= 8

2013-10-20 16:12:45 補充：
1. 是一定要假設f(x) = ax^2 + bx + c 嗎？ax+b之類的形式不可？
若 f(x) = ax+b, 則 f(x+1) - f(x) = (ax+a+b) - (ax+b) = a, 沒有 x 項.

3. 怎樣從2ax + (a+b) = 2x + 1和a = 1, b = 0 得出 f(x) = x^2 + c？
比較 x 項及常數項，所以 2a = 2 及 a+b = 1, 即 a=1, b=0.
因為 f(x) = ax^2 + bx + c, 所以 f(x) = x^2 + c

2013-10-20 16:14:42 補充：
是的，Masterijk 博士，我走漏了眼。多謝指點。

教你簡單一點的方法！！！

If f(x+1)-f(x)=2x+1,
find the value of f(3)-f(1).

代 x = 2 得 f(3) - f(2) = 5 ...[1]
代 x = 1 得 f(2) - f(1) = 3 ...[2]

[1] + [2] 得

f(3) - f(1) = 5 + 3 = 8