A Problem With Functions #algebra #function

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[SidneiMV]

f(3) - f(1) = 1
f(5) - f(3) = 4
f(7) - f(5) = 9

*f(7) - f(1) = 14*

[MrGeorge]

f(x) = (x³ - x + c ) / 24 where c is some constant

[maxhagenauer]

f(x) = x^3/24 - x/24

f(7) - f(1) = 14

[nabilmellouki]

F(2x+1)-F(2x-1)=x^2
Y = 2x+1 => x = (Y-1)/2
F(Y)-F(Y-2)=(Y-1)^2/4
F(Y-2) -F(Y-4) = (Y-3)^2/4
F(Y-4)-F(Y-6) = (Y-5)^2/4
F(Y)-F(Y-6)=((Y-1)^2 + (Y-3)^2 + (Y-5)^2)/4
Y=7
F(7)-F(1)= (36+16+4)/4
F(7)-F(1) = 56/4
F(7)-F(1) = 14

[paulortega]

I took the harder road with difference equations...
let u = 2x - 1, x = (u + 1)/2
f(u+2) - f(u) = (u^2 + 2u + 1)/4 = [u(u-2) + 4u + 1]/4
note if g(u+2) - g(u) = u(u-2)(u-4).. (u-2n-2) then g(u) =
f(u) = [u(u-2)(u-4)/6 + 4u(u-2)/4 + u/2]/4 +f(0)
f(u) = (u^3 - u)/24 + f(0)
f(u) = (u+1)u(u-1)/24 + f(0)
f(1) = f(0)
f(7) = 14 + f(0)
f(7) - f(1) = 14

[pankajchavda]

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