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Write an equation for 1 + 11 + 111 + 1111 ...?

f(n) = 1 + 11 + 111 + 1111 + ... where n is the number of 1's in the last term.

Write an equation for f(n) that doesn't use Epsilon or a series of terms. in your answer.

4 Answers

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  • Hemant
    Lv 7
    10 years ago
    Favorite Answer

    ............ 1 + 11 + 111 + ... to (n) terms

    = (1/9) [ 9 + 99 + 999 + ... to (n) terms ]

    = (1/9) [ (10¹ - 1) + (10² - 1) + (10³ - 1) + ... + (10ⁿ - 1) ]

    = (1/9) [ ( 10¹ + 10² + 10³ + ... + 10ⁿ ) - ( 1 + 1 + 1 + ... n times ) ]

    = (1/9) { (10) [ ( 10ⁿ - 1 ) / (10-1) ] - n }

    = (1/9) { (10/9) [ ( 10ⁿ - 1 ) ] - n } ........................ Ans.

    ______________________________

  • Anonymous
    10 years ago

    n=1

  • 10 years ago

    each term is

    [(10^n)-1]/9

    sum of n-1 terms plus nth term gives sum nth term so

    s(n-1)+[(10^n)-1]/9=s(n)

    likewise we know if we multiply sum n-1 times 10 and and n we get sum of the n terms so

    10[s(n-1)]+n=s(n)

    setting the two equal and solving for s(n-1)

    10[s(n-1)]+n=s(n-1)+[(10^n)-1]/9

    9[s(n-1)]={[(10^n)-1]/9}-n

    9[s(n-1)]={[(10^n)-1]/9}-9n/9

    9[s(n-1)]=[(10^n)-9n-1]/9

    s(n-1)=[(10^n)-9n-1]/81

    ...

    s(n)={10[(10^n)-9n-1]/81}+n

  • Anonymous
    10 years ago

    No, you do it! That ain't my ******* homework.

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