Problem solving an induction problem

Question

Show that for every $n\ge1$ and for every real number $x\ne 1$ $\dfrac {x^{n+1}-1}{x-1}=1+x+x^2+\cdots +x^n$ is valid.

By induction, put $n=1$ $x=2$ and you get for the base case:

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

Then consider $n=m$

$\dfrac {x^{m+1}-1}{x-1}=1+x+x^2+\cdots +x^m$ and assume that holds.

We then calculate $m=n+1$:

$\dfrac {x^{n+2}-1}{x-1}=1+x+x^2+\cdots +x^n+x^{n+1}$ I get:

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

Multiply by $x-1$ and get:

$$x^{n+2}-1=(x-1)(1+x+x^2+\cdots+x^n+x^{n+1})$$

But how do we get further here?

The dots mean that you only go up to the $n$-th power. For $n = 1$ you ignore the $x^2$ and everything ""after"" for example. — Bruno B, Jun 19 '23 at 09:17
The base case ($n=1$) is not what you wrote. The lhs should be $\frac{2^2-1}{2-1}=3$ and the rhs should be $1+2^1=3.$ — Anne Bauval, Jun 19 '23 at 09:17
No, $1=1$ is the case $n=0,$ which is off topic in this exercise. — Anne Bauval, Jun 19 '23 at 09:20
Luthier, for $n=1$ you have $\frac{x^2-1}{x-1}=1+\cdots +x^n=1+x^1$. For $x=2$ this yields $3=3$. — Dietrich Burde, Jun 19 '23 at 09:20
Hint for the proof: just see what happens when you calculate $(x - 1) \sum_{k = 0}^n x^k$, there is no need for induction. — Bruno B, Jun 19 '23 at 09:24

IraeVid · Accepted Answer · 2023-07-04T09:29:24.143

4

When $n=1$ and $x=2$, the equation becomes $$\dfrac{2^2-1}{2-1}=\dfrac{3}{1}=3=1+2$$ so the statement holds true.

Full proof by induction:

We shall induct on $n$. In this proof, $x$ can be any real number except for $1$.

Base case: When $n=1$, we have $\dfrac{x^2-1}{x-1}=1+x$, which is clearly true.

Induction hypothesis: Suppose that when $n=k$, we have $\dfrac{x^{k+1}-1}{x-1}=1+x+x^2+\dots+x^k$.

Inductive step: Consider the case where $n=k+1$.We claim that $\dfrac{x^{k+2}-1}{x-1}=1+x+x^2+\dots+x^{k+1}$.

By the induction hypothesis, $$\begin{align} 1+x+x^2+\dots+x^{k+1} &=\dfrac{x^{k+1}-1}{x-1}+x^{k+1} \\ &=\dfrac{x^{k+1}-1}{x-1}+\dfrac{x^{k+2}-x^{k+1}}{x-1} \\ &=\boxed{\dfrac{x^{k+2}-1}{x-1}} \end{align}$$ This completes our induction.

Hence, for every $n\ge1$ and for every real number $x\ne 1$, $$\dfrac {x^{n+1}-1}{x-1}=1+x+x^2+\cdots +x^n$$

Alternative and more elegant proof:

Let $a=1+x+x^2+\dots+x^{n}$.

Then, $xa=x+x^2+\dots+x^{n}+x^{n+1}$.

Subtracting, we have $(x-1)a=x^{n+1}-1$.

Finally, dividing both sides by $x-1$, we get that $a=\boxed{\dfrac{x^{n+1}-1}{x-1}}$. Isn't that elegant?

edited Jul 04 '23 at 09:29

answered Jun 19 '23 at 09:38

IraeVid

3,216

2

Induction is quite overkill, but clean proof still. – Bruno B Jun 19 '23 at 09:40
2

Well, OP asked for induction proof, but I have edited my post to include an alternative proof. – IraeVid Jun 19 '23 at 09:43
Just checking, what is the intention of the first part with $n = 1, x = 2$? – Calvin Lin Jun 22 '23 at 03:58
At first, OP incorrectly said that when $n=1$ and $x=2$, the formula does not hold and asked why, so I included the explanation that the formula still holds. – IraeVid Jun 22 '23 at 04:29

Problem solving an induction problem

1 Answers1

Full proof by induction:

Alternative and more elegant proof: