Find the values of x and sum of series (as a function of x) for the geometric series which converges of $\sum_{n=0}^{\infty} (-\frac{1}{3})^n(x-7)^n$ Find the values of x.
$\sum_{n=0}^{\infty} (-\frac{1}{3})^n(x-7)^n$
I'm not sure how to combine like terms in this case but I got this:
$(-\frac{x}{3}+\frac{7}{3})^{2n}$
Then separated, x as  -1 < x < 1
I get: -10 < x < -4/3
But this isn't correct. Does anyone know how to solve this and may kindly show me how?
Edit:
Thanks for clearing that confusion up, Dr. Zafar. All I had to do was rearrange the fraction, to get $(\frac{7-x}{3})^n$ and then simplifying it to get $10 > x > 4$.

You were so close!!
Instead of saying $(\frac{-1}{3})^{n}(x-7)^n=(-\frac{x}{3}+\frac{7}{3})^{2n}$
You should have said $(\frac{-1}{3})^{n}(x-7)^n=(-\frac{x}{3}+\frac{7}{3})^{n}$ 
This is a property of exponents such that $a^n*b^n=(ab)^n$ 
The inverse of this property is useful when finding the prime factorization of perfect powers.
$$2304=48^2=(16*3)^2=16^2*3^2=2^8*3^2$$
Can you carry on from here?
