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Thread: 11111...1

  1. #1
    Junior Member Singular's Avatar
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    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 ?
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  2. #2
    Super Member

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    Hello, Singular!

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

    Answer

    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!

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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by Singular View Post
    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
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