It is give function `f(x)` for all real number of `x` such that: `x(2x+1)f(x)+f(1/x)=x+1`

It is give function `f(x)` for all real number of `x` such that:
`x(2x+1)f(x)+f(1/x)=x+1`.

Find the sum of: `S=f(1)+f(2)+f(3)+.......+f(2022)`

Solution

From `x(2x+1)f(x)+f(1/x)=x+1` (1)then we replace `x=1/x` so will get new relation :
            `1/x(2/x+1)f(1/x)+f(x)=1/x+1`

            `f(1/x)+(x^2/(x+2))f(x)=(x^2+x)/(x+2)` (2)

From equation (1)-(2) side to side, then:
            `[x(2x+1)-x^2/(x+2)]f(x)=x+1-(x^2+x)/(x+2)`

            `2x(x^2+1)/(x+2)f(x)=(2x+2)/(x+2)`

Therefore: `f(x)=1/(x(x+1))=1/x-1/(x+1)`

Then, `\sum_{i=1}^{2022}f(k)`= `\sum_{i=1}^{2022}(1/k-1/(k+1))` = `1-1/2022=2021/2022`

Hence, the sum of: `S=f(1)+f(2)+f(3)+.......+f(2022)=2021/2022`

Solution by: Thin Sokkean