Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm trying to do my Maths assignment but I can't get this done. I can do other questions but this one is different. This is a picture of the question.

pic

My usual first step is to proof it when n = 1. But the problem is, that I don't know what x is. It says x is not equal to 1, but I still don't know what to do next.

Thank you.

share|improve this question
    
The result need to be independent of what $x$ is (except of course $x \ne 1$). –  Peter Grill Oct 31 '12 at 0:41
    
No matter what $x$ is, when $n=1$ the left side is $1+x$ and the right side is $(1-x^2)/(1-x)$, right? –  Gerry Myerson Oct 31 '12 at 0:43
    
@GerryMyerson How's the left side 1 + x? x ^i = x ^ 0 = 1. Right? –  Adegoke A Oct 31 '12 at 0:46
1  
Yes, so the left side is $x^0+x^1$, which is $1+x$. –  Gerry Myerson Oct 31 '12 at 0:55
add comment

4 Answers

up vote 2 down vote accepted

Induction involving $x$ is no different from other types of induction. Just treat $x$ normally and we can still prove it for the general case.

Let $$P_n:=\sum_{i=0}^nx^i=\frac{1-x^{n+1}}{1-x}$$

Then for the base case i.e. $n=0$ we have $$\sum_{i=0}^0x^i=\frac{1-x^{1+0}}{1-x}$$ $$1=\frac{1-x}{1-x}$$ $$1=1$$

He have esablished the RHS=LHS and so $P_0$ is true.

Now assume that the statement is true for n. $$\sum_{i=0}^nx^i=\frac{1-x^{n+1}}{1-x}$$

We must show that it is also true for $n+1$.

For the RHS we have $$RHS=\frac{1-x^{(n+1)+1}}{1-x}=\frac{1-x^{n+2}}{1-x}$$

And for the LHS we have $$LHS=\sum_{i=0}^{n+1}x^i=\sum_{i=0}^{n}x^i+x^{n+1}=\frac{1-x^{n+1}}{1-x}+x^{n+1}=\frac{1-x^{n+1}}{1-x}+\frac{x^{n+1}(1-x)}{(1-x)}$$ $$=\frac{1-x^{n+1}+x^{n+1}(1-x)}{1-x}=\frac{1-x^{n+1}+x^{n+1}-x^{n+2}}{1-x}=\frac{1-x^{n+2}}{1-x}$$ Since LHS=RHS $P_{n+1}$ is true. The rest follows.

share|improve this answer
1  
Did you notice that this was an assignment problem? Did you not want to leave anything for OP to do? –  Gerry Myerson Oct 31 '12 at 6:00
add comment

It doesn't matter what $x$ is, as long as it isn't $1$ (so $1-x$ isn't $0$, and you can divide by it). Just treat it as a normal inductive proof:

Suppose $\sum_{i=0}^{n}{x^{i}}=\frac{1-x^{n+1}}{1-x}$. Then

$$\sum_{i=0}^{n+1}{x^{i}}=\sum_{i=0}^{n}{x^{i}}+x^{n+1}=\frac{1-x^{n+1}}{1-x}+x^{n+1}=\frac{(1-x^{n+1})+(1-x)x^{n+1}}{1-x}$$

$$=\frac{1-x^{n+2}}{1-x},$$

which completes the inductive step.

The base case $n=0$ is LHS=$x^{0}+x^{1}=1+x$ (no matter what $x$ is), RHS=$\frac{1-x^{2}}{1-x}=1+x$, so it's true for all $n$.

share|improve this answer
add comment

For $n=1$, the right hand side is

$$\begin{align*} \frac{1 - x^2}{1-x} &= \frac{(1-x)(1+x)}{1-x} \\ &= 1+x, \quad x \ne 1 \end{align*}$$

which is identical to the right hand side.

share|improve this answer
    
But what about x^1? x^i =1. 1 does not equal to 1 + x. –  Adegoke A Oct 31 '12 at 0:50
    
The left hand side for $n=1$ expands to $x^0+x^1 = 1 + x, x \ne 0$ –  Peter Grill Oct 31 '12 at 0:53
add comment

You need to let $x$ be an arbitrary number, which is just not $1$.

For the base case ($n = 0$): We have $$x^0 = 1 = \frac{1-x}{1-x} = \frac{1-x^{0+1}}{1-x}$$ So the theorem holds in the base case.

For the inductive step: $$\sum_{i=0}^{n+1} x^i = \sum_{i=0}^{n} x^i + x^{n+1}$$

And by induction we know $\sum_{i=0}^n x^i = \frac{1 - x^{n+1}}{1-x}$, so plugging this in we get $$\sum_{i=0}^{n+1}x^i = \frac{1-x^{n+1}}{1-x} + x^{n+1} = \frac{1-x^{n+1}}{1-x} + \frac{(1-x)x^{n+1}}{1-x} = \frac{1 - x^{n+2}}{1-x}.$$

So by induction we're done.

One can also note that there is another proof, without induction that is neater.

$$(\sum_{i=0}^{n} x^i) (1-x) = \sum_{i=0}^{n} x^i - \sum_{i=0}^{n} x^{i+1} = 1 - x^{n+1}$$ because all the terms in the middle cancel out.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.