tan⁡x+12tan⁡x2+122tan⁡x22+…+12n−1tan⁡x2n−1 is equal to

We can write tan x=tan2⁡x−1+1tan⁡x=cot⁡x−2cot⁡2x so tan⁡x+12tan⁡x2+122tan⁡x22+⋯+12n−1tan⁡x2n−1 =cot⁡x−2cot⁡2x+12cot⁡x2−2cot⁡x+122cot⁡x22−2cot⁡x2+⋯+12n−1cot⁡x2n−1−2cot⁡x2n−2  =12n−1cot⁡x2n−1−2cot⁡ 2x