I have no idea how to prove this

Question

I am asked to prove the equation:

$$\frac{10^n-1}9 = \underbrace{111\cdots1}_{n\space\text{digits}}$$ where $n$ is a positive integer. Can anyone help me to prove it?

Answer 1

$\frac {10^n - 1}9 = \underbrace{1111.....111}_{n\text{ times}} \iff$

$10^n - 1 = 9\times \underbrace{1111.....111}_{n\text{ times}}\iff$

$10^n -1 = \underbrace{99999.....99999}_{n\text{ times}}\iff$ $10^n = \underbrace{99999.....99999}+1$

Perhaps more sophisticated is noticing $(a-1)(a^{n-1} + a^{n-2} + .....+a + 1)=$

$(a^{n} + a^{n-1} .....+a^2 + a) - (a^{n-1} + ..... + a + 1) = a^n - 1$.

Just replace $a = 10$ and ... what do you get?

Answer 2

You may simply it first. Think about it this way too. Let $n$ be 3.

$$10^3 - 1=1000 - 1 = 999$$ $$\frac{999}{9} = 111$$

Now generalize for any $n$.

By the way, this is not a formal proof.

Added

Now, you may consider this as a kind of formal proof. Since $10^n = 100..0$ having n zeros. Subtracting 1 from this number yeilding n nines, dividing by 9 yielding n ones.

Answer 3

Note that: $$\require{cancel} \frac{10^n-1}9=\frac{\cancel{(10-1)}(10^{n-1}+10^{n-2}+\cdots +10^1+10^0)}{\cancel{9}}=\\ \underbrace{100...00}_{n}+\underbrace{100...00}_{n-1}+\cdots +\underbrace{10}_{2}+\underbrace{1}_{1}=\underbrace{111...111}_{n}.$$

Answer 4

Induction :

$\dfrac{10^1-9}{9}= 1$

Assume the formula valid for $n$.

Step $n+1:$

$\dfrac{10^{n+1}-1}{9}=$

$\dfrac{(9+1)10^n -1}{9}=$

$\dfrac{9 \cdot 10^n +(10^n-1)}{9}=$

$10^n + 1111111...(n $times$)=$ (hypothesis)

$1111111...((n+1)$ times$)).$

Note;

$10^1 = 10$ $(1$ appears as the $2$nd digit)

$10^2 = 100 (1$ appears as the $3$rd digit)

$10^n : 1$ appears as the $(n+1)$st digit .