Proving this sequence ( probably using induction? ) Question: 
For $n>0$ and $\theta\in\Re$, Let $$t_n = \sum_{k=1}^n (-1)^k 3^{k-1} cos^3{\frac{\theta}{3^k}}$$
Using the identity $cos^3\phi=\frac{1}{4}cos3\phi+\frac{3}{4}cos\phi$,
Show that $$t_n = \frac{(-1)^n 3^n}{4}cos\frac{\theta}{3^n} - \frac{1}{4}cos\theta$$ and evaluate$ \lim_\limits{n \to \infty} t_n$
My attempt(using induction):
Base case(n=1) : $$t_1 = \sum_{k=1}^1 (-1)^k 3^{k-1} cos^3{\frac{\theta}{3^k}}$$
$$=(-1)3^{1-1}cos^3\frac{\theta}{3}$$
$$=-(\frac{1}{4}cos\theta + \frac{3}{4}cos\frac{\theta}{3})$$
$$=\frac{(-1)^1 3^1}{4}cos\frac{\theta}{3^1}-\frac{1}{4}cos\theta$$
Assumption: 
$$t_m = \frac{(-1)^m 3^m}{4}cos\frac{\theta}{3^m} - \frac{1}{4}cos\theta$$
$$t_{m+1} = \sum_{k=1}^{m+1} (-1)^k 3^{k-1} cos^3{\frac{\theta}{3^k}}$$
$$=(-1)^{m+1}3^mcos^3\frac{\theta}{3^{m+1}}+ \sum_{k=1}^m (-1)^k 3^{k-1} cos^3{\frac{\theta}{3^k}}$$
$$=(-1)^{m+1}3^mcos^3\frac{\theta}{3^{m+1}}+ t_m$$
Then using assumption , am I correct?

Hint. One may observe that, for $k=1,2,3\cdots,$ we have (why?)
$$
\frac{(-1)^k 3^k}{4}\cdot\cos\frac{\theta}{3^k}-\frac{(-1)^{k-1} 3^{k-1}}{4}\cdot\cos\frac{\theta}{3^{k-1}}=(-1)^k 3^{k-1} \cdot\cos^3{\frac{\theta}{3^k}}
$$ then summing from $k=1$ to $k=n$ one may recognize a telescoping sum.
