Question

F(x) = g(x+1)

f(y) = 2018y + 2016y +. . . . .+ 2y

g(x) = h(2x)-x

If h(2018) =
`a^(3)` and a is an integer, then a = ?

Answer 1

`g(x)=h(2x)-x\implies h(2x)=g(x)+x`

`f(x)=g(x+1)\implies g(x)=f(x-1)`

Taken together, we get

`h(2x)=f(x-1)+x`

Since
`2018=2\cdot1009`, the above gives

`a^3=f(1008)+1009`

We have

`f(y)=\displaystyle\sum_(i=0)^(1008)(2018-2i)y=1019090y`

so that

`a^3=1019090\cdot1008+1009`

`a^3=1027243729=1009^3`

`\implies a=1009`