1. ## 11111...1

Prove that

$
\underbrace {11111.......1}_{n1's} = \frac{{10^n - 1}}{9};For{\rm }n \ge 1
$

ans:

for n=k ---> $
\underbrace {11111.......1}_{k1's} = \frac{{10^k - 1}}{9}
$

for n=k+1-----> $
10(\underbrace {11....1}_{k1's}) + 1 = 10(\frac{{10^k - 1}}{9}) + 1 = \frac{{10^{k + 1} - 10}}{9} + \frac{9}{9} = \frac{{10^{k + 1} - 1}}{9}
$

Is it right ?

2. Hello, Singular!

Prove that: . $\underbrace {111 \cdots1}_{n\,1's} \;= \;\frac{10^n-1}{9},\;\text{for }n \ge 1$

for $n = k\!:\;\underbrace{11111 \cdots1}_{k\,1's} \;= \;\frac{10^k-1}{9}$

for $n=k+1\!:\;10(\underbrace {111\cdots1}_{k\,1's}) + 1 \;= \;10\left(\frac{10^k-1}{9}\right) + 1 \;=$ $\frac{10^{k+1}-10}{9} + \frac{9}{9} \;=\; \frac{10^{k+1}-1}{9}$

Is it right ?

Looks good to me . . . nice work!

3. Originally Posted by Singular
Prove that

$
\underbrace {11111.......1}_{n1's} = \frac{{10^n - 1}}{9};For{\rm }n \ge 1
$
For all $n \ge 1$ the lefthand side is:

$
\sum_0^{n-1} 10^n
$

which is a geometric series with sum: $(1-10^n)/(1-10)=(10^n-1)/9$

RonL