$$ \frac{x^{n+1}-1}{x-1} = 1 + x + x^2 + \dots + x^n $$ where $x\neq 1$, $x\in \mathbb R$. I am really tired to prove that questions. I can not understand any one. Please help me.....How to prove that, for each $n\in \mathbb N$ (using Mathematical induction)
My try
how to complete this steps?
Prove that the equality is true when $n=0$ by direct substitution.
Then write the right-hand side with $n+1$ in place of $n$ and use the equality for the first $n$ terms.
This is equivalent to $(x-1)(1+x+\cdots x^n) = x^{n+1}-1$
Base case, $n=0$,
$$(x-1)(x^0)=x^{0+1}-1$$
Now assume it is true up to $n-1$, so
$$(x-1)(1+x+\cdots x^{n-1}+x^n) = [(x-1)(1+x+\cdots x^{n-1})] + (x-1)(x^n) $$ $$= [x^n - 1] + [x^{n+1}-x^n] =x^{n+1}-1 $$
Now divide the $(x-1)$ out to RHS
$$(1+x+\cdots x^{n-1}+x^n) = \frac{x^{n+1}-1}{x-1}$$
which we can do so long as $x\neq 1$.
First of all you verify that this hold for some number, i.e. $n=0$. Then you assume that it holds for some $n=k$. And then you need to prove the result for $n=k+1$ by using that assumption. Now, $\frac{x^{k+1}-1}{x-1}=x^k+\frac{x^k-1}{x-1}$. From the Induction Hypothesis for $n=k$ we have that $\frac{x^{k}-1}{x-1}=1+x+x^2+...+x^{k-1}$. Hence, $\frac{x^{k+1}-1}{x-1}=x^k+\frac{x^k-1}{x-1}=x^k+(1+x+x^2+...+x^{k-1})$. So it holds for $n=k+1$ also and this complete the Induction. Thus $\frac{x^{n+1}-1}{x-1}=1+x+x^2+...+x^n$ holds for any $n$.
Assume it holds for $n-1$. then $1+x+x^2\dots+x^{n-1}=\dfrac{x^{n}-1}{x-1}$. From this we get:
$1+x+x^2\dots+x^{n-1}+x^n=\dfrac{x^{n}-1}{x-1}+x^n=\dfrac{x^n-1+x^n(x-1)}{x-1}=\dfrac{x^n-1+x^{n+1}-x^n}{x-1}=\dfrac{x^{n+1}-1}{x-1}$
Meaning it holds for $n$
By direct substitution, we know that this is true for n = 1.
Let this be true n = k
then
$1 + x^2 + .... + x^k = \frac{x^{k+1} - 1}{x-1}$
Now for check whether it is true for n = k+1
$1 + x^2 + .... + x^k + x^{k+1} = \frac{x^{k+1} - 1}{x-1} + x^{k+1}$
= $\frac{x^{k+2} - 1}{x-1}$
Thus this is true for n = k + 1 also.
Thus we can say that this equation will be true for all natural numbers.
