0

ANS: Not sure how to do this. I wrote Induction: Prove P(K) = P(k+1) so assume P(k) is true. Statement = P(k+1)

1 + a + a^2 + a^3 + .... a^n-1 = 1-a^n/1-a = a + a + a^2 + a^3... a^k-1 = 1-a^k/1-a

a + a + a^2 + a^3... a^k-1 + k + 1 = 1-a^k(k+1)/1-a

?

I think I did this completely wrong

Lostsoulaside
  • 59
  • 4
    Your question is unreadable. Please use MathJax to improve the readability. https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – 5xum Oct 12 '20 at 09:50

2 Answers2

1

To prove: $1 + a + a^2 ... + a^{n-1} = \frac{1-a^n}{1-a}$, for $n \in \mathbb{N}$.

Let $P(n)$ denote $1 + a + a^2 ... + a^{n-1} = \frac{1-a^n}{1-a}$ for $n \in \mathbb{N}$. We shall use weak induction, i.e. show $P(1)$ is true, assume for $P(k)$, show for $P(k+1)$.

The base case, i.e. $P(1)$ is easy to see, since the LHS $= 1$ and RHS $= \frac{1-a}{1-a} = 1$.

Now, assume $P(k)$ to be true for some $n \in \mathbb{N}$. Let us show that $P(k+1)$ is true.

$$1 + a + a^2 ... + a^{k-1} + a^k = \frac{1-a^k}{1-a} + a^k = \frac{1-a^{k+1}}{1-a}$$ which shows that $P(k+1)$ is true, and completes the proof. Note that the first equality follows from the fact that $P(k)$ is true (assumed right before).

stoic-santiago
  • 13,705
0

You don't really need a formal induction here: the formula is equivalent to $$(1-a)(1+a+a^2+\dots+a^{n-1})=1-a^n, $$ a high-school factorisation formula, that you can prove doing the multiplication in the l.h.s.: $$\begin{alignedat}{6}1&+{}&a&+{}&a^2&+{}&\dotsm\dotsm&+{}&a^{n-1} \\ &-{}&a&-{}&a^2&-&a^3-\dotsm&-{}&a^{n-1}&-a^{n}\\\hline =&&&&&&1-a^n \end{alignedat}$$

Bernard
  • 175,478