1. ## 11111...1

Prove that

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

ans:

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

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

Is it right ?

2. Hello, Singular!

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

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

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

Is it right ?

Looks good to me . . . nice work!

3. Originally Posted by Singular
Prove that

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

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

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

RonL