What is the mathematical formula for nth term of 7 + 77 + 777 + ...+ n?

Don't tell me "n 7s" or something like that. That's not a math formula.

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  • Ian H
    Lv 7
    9 months ago

    It helps to think what would be the nth term of 9 + 99 + 999 + ...+ n?

    As an example the third term, 999 is 10^3 – 1 and they are all like that.

    The general nth term is (7/9)( 10^n – 1)

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  • sepia
    Lv 7
    9 months ago

    7, 77, 777, 7777, ....

    an = 7/9 (10^n - 1)

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  • 9 months ago

    Let S(n)=7+77+777+...+the nth term be

    7[1+11+111+1111+...+T(n)]

    Consider the sequence

    ........1, 11, 111, 1111,.....,T(n)

    (d): ..10,.100,.1000,.......,T(n)

    where d(n-1)=T(n)-T(n-1) & n=2,3,...

    =>

    T(n)=10^(n-1)+T(n-1)

    =>

    T(2)=10+T(1)

    T(3)=10^2+10+T(1)

    T(4)=10^3+10^2+10+T(1)

    --------

    T(n)=10^(n-1)+10^(n-2)+..+10+

    T(1)

    =>

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

    T(n)=(10^n-1)/9

    [T(1)=1]

    =>

    The nth term of

    S(n)=7T(n)=7[10^n-1]/9

    Check:

    The 2nd term=7[100-1]/9=77

    The 3rd term=7[1000-1]/9=777

    The 4th term=7[10000-1]/9=7777

    etc.

    valid.

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  • 9 months ago

    What is the mathematical formula for nth term of 7 + 77 + 777 + ...+ n?

    7n + 70(n-1) + 700(n-2) + ... + 7*10^(n-1) (1)

    7((n)10^0  + (n-1)10^1 + (n-2)10^2 + ... + (1)10^(n-1))

    7( [k = 1 to n] ∑ (n+1-k)10^(k-1) )

    7(10^(n+1) - 9n - 10)/81 ← ← ← ANSWER

    e.g.

    T(1) = 7(10^(1+1) - 9*1 - 10)/81 = 7(81)/81 = 7

    T(2) = 7(10^(2+1) - 9*2 - 10)/81 = 7(972)/81 = 84 = 7 + 77

    T(3) = 7(10^(3+1) - 9*3 - 10)/81 = 7(10^(3+1) - 9*3 - 10)/81 = 7(9963)/81 = 861 = 7 + 77 + 777

    etc.

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  • 9 months ago

    Here's the function that will return the nth term of that sequence.

    a[n] = 7*(10^n - 1)/9

    Let's double-check it:

    a[1] = 7*(10^1 - 1)/9 = 7 * 9/9 = 7

    a[2] = 7*(10^2 - 1)/9 = 7 * 99/9 = 77

    a[3] = 7*(10^3 - 1)/9 = 7 * 999/9 = 777

    etc.

    Answer:

    a[n] = 7*(10^n - 1)/9

    P.S. This is not the partial sum of the first n terms; it's the nth term. Ask a different question if you are looking for the nth partial sum.

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  • Anonymous
    9 months ago

    7*sum(10^x) for x=0:N

    You never said we needed to reduce.

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  • 9 months ago

    7/9 (10^n - 1) ..................

    For example 5th term

    7/9 (100000 -1) = 7/9 * 99999 = 77777

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  • 9 months ago

    Each term multiplies the previous term by 10 and adds 7.

    so:

    f(n) = f(n - 1) * 10 + 7 when n > 1

    f(n) = 7 when n = 1

    Then it's an iterative process to work out say f(4), so it looks like:

    f(n) = f(n - 1) * 10 + 7

    f(4) = f(3) * 10 + 7

    f(3) = f(2) * 10 + 7

    f(2) = f(1) * 10 + 7

    f(2) = 7 * 10 + 7

    Then it unwinds:

    f(2) = 70 + 7

    f(2) = 77

    f(3) = f(2) * 10 + 7

    f(3) = 77 * 10 + 7

    f(3) = 770 + 7

    f(3) = 777

    f(4) = f(3) * 10 + 7

    f(4) = 777 * 10 + 7

    f(4) = 7770 + 7

    f(4) = 7777

    • Anon E9 months agoReport

      I was hoping for a single formula. Maybe there isn't one. 

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  • TomV
    Lv 7
    9 months ago

    In base 8 (octal) arithmetic, it would be (10₈^n -1)

    • TomV
      Lv 7
      9 months agoReport

      Free with the insults aren't you, sport, but a little short on refutation. Is that because you don't understand any number systems other than decimal? Comment reported by the way.

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  • Mark
    Lv 7
    9 months ago

    "7+77+777+... +N" IS a formula.

    • TomV
      Lv 7
      9 months agoReport

      But it is not the formula for the nth term.

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