Answer: (a) \(1+w+w^2+\cdots+w^{n-1}=\dfrac{w^n-1}{w-1}\), for \(w\neq 1\).

(b) \((1+\mathrm{i}\tan\theta)^k=\sec^k\theta\,(\cos k\theta+\mathrm{i}\sin k\theta)\).

(c) \(\displaystyle \sum_{k=0}^{n-1}\sec^k\theta\sin k\theta=\cot\theta\left(1-\sec^n\theta\cos n\theta\right)\).

(d) \(\displaystyle \sum_{k=0}^{6m-1}2^k\sin\left(\frac{k\pi}{3}\right)=\frac{\sqrt3}{3}\left(1-2^{6m}\right)\).

(a) This is a geometric series with first term \(1\), common ratio \(w\), and \(n\) terms, so

\(1+w+w^2+\cdots+w^{n-1}=\dfrac{w^n-1}{w-1}\), provided \(w\neq 1\).

(b) Rewrite

\(1+\mathrm{i}\tan\theta=1+\mathrm{i}\dfrac{\sin\theta}{\cos\theta}=\dfrac{\cos\theta+\mathrm{i}\sin\theta}{\cos\theta}=\sec\theta(\cos\theta+\mathrm{i}\sin\theta).\)

Therefore

\((1+\mathrm{i}\tan\theta)^k=\left[\sec\theta(\cos\theta+\mathrm{i}\sin\theta)\right]^k=\sec^k\theta(\cos\theta+\mathrm{i}\sin\theta)^k.\)

Using de Moivre's theorem,

\((\cos\theta+\mathrm{i}\sin\theta)^k=\cos k\theta+\mathrm{i}\sin k\theta.\)

Hence

\((1+\mathrm{i}\tan\theta)^k=\sec^k\theta(\cos k\theta+\mathrm{i}\sin k\theta).\)

(c) Consider

\(\displaystyle \sum_{k=0}^{n-1}(1+\mathrm{i}\tan\theta)^k.\)

This is a geometric series with common ratio \(1+\mathrm{i}\tan\theta\), so

\(\displaystyle \sum_{k=0}^{n-1}(1+\mathrm{i}\tan\theta)^k=\frac{(1+\mathrm{i}\tan\theta)^n-1}{\mathrm{i}\tan\theta}.\)

Using part (b),

\((1+\mathrm{i}\tan\theta)^n=\sec^n\theta(\cos n\theta+\mathrm{i}\sin n\theta).\)

So

\(\displaystyle \sum_{k=0}^{n-1}(1+\mathrm{i}\tan\theta)^k=\frac{\sec^n\theta(\cos n\theta+\mathrm{i}\sin n\theta)-1}{\mathrm{i}\tan\theta}.\)

Now \(\dfrac{1}{\mathrm{i}\tan\theta}=-\mathrm{i}\cot\theta\), hence

\(\displaystyle \sum_{k=0}^{n-1}(1+\mathrm{i}\tan\theta)^k=-\mathrm{i}\cot\theta\left[\sec^n\theta(\cos n\theta+\mathrm{i}\sin n\theta)-1\right].\)

Expanding,

\(\displaystyle \sum_{k=0}^{n-1}(1+\mathrm{i}\tan\theta)^k=-\cot\theta\sec^n\theta(\mathrm{i}\cos n\theta-\sin n\theta)+\mathrm{i}\cot\theta.\)

Also, from part (b),

\((1+\mathrm{i}\tan\theta)^k=\sec^k\theta\cos k\theta+\mathrm{i}\sec^k\theta\sin k\theta,\)

so the imaginary part of the left-hand side is

\(\displaystyle \sum_{k=0}^{n-1}\sec^k\theta\sin k\theta.\)

The imaginary part of the right-hand side is

\(\cot\theta-\cot\theta\sec^n\theta\cos n\theta=\cot\theta\left(1-\sec^n\theta\cos n\theta\right).\)

Therefore

\(\displaystyle \sum_{k=0}^{n-1}\sec^k\theta\sin k\theta=\cot\theta\left(1-\sec^n\theta\cos n\theta\right).\)

(d) Take \(\theta=\dfrac{\pi}{3}\) and \(n=6m\) in the result from part (c). Then

\(\sec\left(\frac{\pi}{3}\right)=2,\qquad \cot\left(\frac{\pi}{3}\right)=\frac{1}{\sqrt3}=\frac{\sqrt3}{3}.\)

So

\(\displaystyle \sum_{k=0}^{6m-1}2^k\sin\left(\frac{k\pi}{3}\right)=\frac{\sqrt3}{3}\left(1-2^{6m}\cos\left(\frac{6m\pi}{3}\right)\right).\)

Since

\(\cos\left(\frac{6m\pi}{3}\right)=\cos(2m\pi)=1,\)

it follows that

\(\displaystyle \sum_{k=0}^{6m-1}2^k\sin\left(\frac{k\pi}{3}\right)=\frac{\sqrt3}{3}\left(1-2^{6m}\right).\)