Find the values of x and sum of series (as a function of x) for the geometric series which converges of $\sum_{n=0}^{\infty} (-\frac{1}{3})^n(x-7)^n$

Question

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 $10 > x > 4$.

Answer 1

$$S=\sum_{n=0}^{\infty}\left(\frac{-1}{3}\right)^n (x-7)^n=\frac{1}{1-\frac{7-x}{3}}=\frac{3}{x-4}.$$ This sum is valid for $$-1< \frac{7-x}{3} <1 \Rightarrow 4<x<10.$$

Answer 2

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?