1. ## Proof question.

How do you prove this?

Many thanks!

2. Hi

$\displaystyle (1-x)\sum\limits_{k=0}^{N}x^k=\sum\limits_{k=0}^{N}x^ k-\sum\limits_{k=1}^{N+1}x^{k}=1-x^{N+1}$

If $\displaystyle x\neq 1$ then $\displaystyle x-1\neq 0$, and you can divide by $\displaystyle 1-x$.

3. Originally Posted by bluebiro
How do you prove this?

Many thanks!
Let $\displaystyle S = 1 + x + x^2 + ... + x^N$

Then $\displaystyle xS =$ $\displaystyle x + x^2 + ....+ x^N + x^{N+1}$

Now subtract the two equations:

$\displaystyle S - xS = (1 + x + x^2+ ....+ x^N) - (x + x^2 + ....+ x^N + x^{N+1})$

Group terms now, so that they cancel,

$\displaystyle (1 - x)S = 1 +(x - x) + (x^2 - x^2)+ ....+(x^N - x^N) - x^{N+1}$

After canceling, we have:

$\displaystyle (1 - x)S = 1 - x^{N+1}$

Now since $\displaystyle x \neq 1$, divide by $\displaystyle (1 - x)$ on both sides and we are done!

NOTE: This is exactly what clic-clac did, but I have explicitly written down the terms so that you can "see" the solution clearly. Try reading through clic-clac's post, to learn elegant ways of writing proofs involving sums.

4. $\displaystyle \sum\limits_{k=0}^{n}{x^{k}}=\frac{1}{1-x}\overbrace{\sum\limits_{k=0}^{n}{\left( x^{k}-x^{k+1} \right)}}^{\text{Telescoping Sum}}=\frac{1-x^{n+1}}{1-x}.\quad\blacksquare$