Question

We have `P(x)=(x^2+x+1)^2015+x+1`. How to calculate `sum_(k=1)^4030 1/x_k`?

Answer 1

`1008`

Explanation:

Calling

`p(x)=(x^2+x+1)^2015+x+1` we have

`p(0)=c_0 = 2`
`p'(0)=c_1/(1!) = 2016`

then

`sum_(k=1)^4030 1/x_k = c_1/c_0 = 1008`