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Math Help - Lost again .

  1. #1
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    Lost again .

    Find the nth term and the sum of the first n terms of the series .

    1+11+111+1111+11111+...

    I just see one more '1' is added . Not sure how to do it ..

    THanks
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  2. #2
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    Quote Originally Posted by thereddevils View Post
    Find the nth term and the sum of the first n terms of the series .

    1+11+111+1111+11111+...

    I just see one more '1' is added . Not sure how to do it ..

    THanks
    Here's a start for you:

    The terms of the series are 1, \, 1 + 10, \, 1 + 10 + 10^2, \, 1 + 10 + 10^2 + 10^3, \, ....
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  3. #3
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    Hello, thereddevils!

    This is not an easy problem . . .


    Find the n^{th} term and the sum of the first n terms of the series:

    . . . . . . 1+11+111+1111+11111+\hdots
    The general term of this sequence is tricky to derive.

    The 5th term has five 1's.
    How can we "create" the number 11,111 from a formula?

    We can get five 9's from: . 10^5-1 \:=\:99,\!999 . . . then divide by 9.

    . . That is: . \frac{10^5-1}{9} \:=\:11,\!111

    \text{In general: }\:\frac{10^n-1}{9} \;=\;\underbrace{111\hdots 111}_{n\:1's}

    Therefore, the n^{th} term is: . \boxed{a_n \:=\:\frac{10^n-1}{9}}


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    The sum of the first n terms is: . S \;=\; \sum^n_{k=1}\frac{10^k-1}{9}

    . . which equals: . \frac{1}{9}\sum^n_{k=1}\left[10^k - 1\right] \;=\;\frac{1}{9}\left[\sum^n_{k=1}10^k - \sum^n_{k=1} 1 \right]


    The first sum is: . \sum^n_{k=1} 10^k \;=\;10 + 10^2 + 10^3 + \hdots + 10^m
    . . . This is a geometric series whose sum is: . \frac{10(10^n-1)}{9}

    The second sum is: . \sum^n_{k=1}1 \:=\:n


    So we have: . S \;=\;\frac{1}{9}\left[\frac{10(10^n-1)}{9} - n\right]


    . . Therefore: . \boxed{S \;=\;\frac{1}{81}\left(10^{n+1} - 9n - 10\right]}

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  4. #4
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    Thanks a lot Soroban .
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