Proving \sum_{k=0}^{n} x^{k} = \frac{1-x^{n+1}}{1-x}: A Step-by-Step Guide

[courtrigrad]

I want to show that $$\sum_{k=0}^{n} x^{k} = \frac{1-x^{n+1}}{1-x}$$ using the additive, homogeneous, and telescoping properties of summation. In a hint it says to write the sum as $$(1-x)\sum_{k=0}^{n} x^{k}$$. How did they arrive at this? Did they factor out the $$1-x$$. I don't see how they did this. I would then write $$x^{k}$$ as $$x^{k+1} - (x-1)^{k+1}$$. Then what?

Thanks

 

[Curious3141]

I want to show that $$\sum_{k=0}^{n} x^{k} = \frac{1-x^{n+1}}{1-x}$$ using the additive, homogeneous, and telescoping properties of summation. In a hint it says to write the sum as $$(1-x)\sum_{k=0}^{n} x^{k}$$. How did they arrive at this? Did they factor out the $$1-x$$. I don't see how they did this. I would then write $$x^{k}$$ as $$x^{k+1} - (x-1)^{k+1}$$. Then what?

Thanks

Let the sum be represented by S. You want to find a neat expression for S.

One way to do that is to multiply S by (1-x). List out a few terms of S (maybe the first four, then "..." then the last two terms). Now multiply that by (1-x) term by term. Add up the like powers (same exponent of x) and see what happens. You will end up very quickly at the required proof.

 

[VietDao29]

Another way is that:
Let:
$$S = \sum_{n = 1} ^ k (x ^ n)$$
Now multiply both sides by x, we have:
$$x S = \sum_{n = 1} ^ k (x ^ {n + 1}) = \sum_{n = 2} ^ {k + 1} (x ^ n)$$
Now, from the 2 equations above, what can you do to get (1 - x) S?
Can you go from here? :)

 

[courtrigrad]

Yup, I got it.

Thanks