How to solve the following infinite sum:
$$\sum_{n=0}^\infty \frac{[(-25)^n+(-9)^n+(-1)^n]x^{(2n)}}{(2n)!}$$
I have no idea where to go from here, but I began with $\cos(5x) + \cos(3x) + \cos(x)$. I want to solve this way, making one unified trig function. Ideas?
You can use complex numbers to turn $\cos x + \cos 3x + \cos 5x$, which follows from the series, into a somewhat more unified trigonometric function. Consider $e^{ix} = \cos x + i \sin x$, and the current expression as $\Re(e^{ix} + e^{3ix} + e^{5ix})$.
This forms a geometric series, with common ratio $e^{2ix}$ and initial term $e^{ix}$:
$$ e^{ix} + e^{3ix} + e^{5ix} = \frac{e^{ix} (1 - e^{6ix})}{1 -e^{2ix}} \\ \because 1-e^{ix} = 1-\cos x - i \sin x = 2 \sin^2 \frac{x}{2}-2i \sin \frac{x}{2} \cos \frac{x}{2} = -2i e^\frac{ix}{2} \sin \frac{x}{2} \\ \therefore e^{ix} + e^{3ix} + e^{5ix} = e^{ix}\times\frac{-2i e^{3ix} \sin 3x }{-2i e^{ix} \sin x } = e^{3ix} \frac{\sin 3x}{\sin x}= \frac{(\cos 3x + i \sin 3x)\sin 3x}{\sin x} \\ $$
Take the real part of both sides.
$$ \therefore \sum_{n=0}^\infty \frac{[(-1)^n + (-9)^n + (-25)^n]x^{2n}}{(2n)!} = \cos x+\cos 3x+\cos 5x= \frac{\cos 3x \sin 3x}{\sin x} = \frac{\sin 6x}{2 \sin x} $$
Hope this helps.