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I have a question about finding the sum formula of n-th terms.

Here's the series:

$5+55+555+5555$+......

What is the general formula to find the sum of n-th terms?

My attempts:

I think I need to separate 5 from this series such that:

$5(1+11+111+1111+....)$

Then, I think I need to make the statement in the parentheses into a easier sum:

$5(1+(10+1)+(100+10+1)+(1000+100+10+1)+.....)$

= $5(1*n+10*(n-1)+100*(n-2)+1000*(n-3)+....)$

Until the last statement, I don't know how to go further. Is there any ideas to find the general solution from this series?

Thanks

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14 Answers 14

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$$5+55+555+5555+\cdots+\overbrace{55\dots5}^{n\text{ fives}}$$ $$=\frac59(9+99+999+9999+\cdots+\overbrace{99\dots9}^{n\text{ nines}})$$ $$=\frac59(10^1-1+10^2-1+10^3-1+\cdots+10^n-1)$$ $$=\frac59(10^1+10^2+10^3+\cdots+10^n-n)$$ $$=\frac59\left(\frac{10^{n+1}-10}{9}-n\right).$$

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    $\begingroup$ @SteveJessop Thanks for the correction. $\endgroup$ – bof Nov 16 '14 at 14:47
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    $\begingroup$ oh, the idea is using the properties of 10^n to make it easier. Thanks for your answer, it's really help! $\endgroup$ – akusaja Nov 17 '14 at 3:11
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    $\begingroup$ @bof Be careful. The way you've notated your first line, by putting $+\cdots,$ you're summing to infinity, rather than summing to $n$ terms. $\endgroup$ – beep-boop Nov 18 '14 at 22:29
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    $\begingroup$ @bof - very well deserved though. a nice and elegant proof without requiring a single $\sum$ sign. :) $\endgroup$ – hypergeometric Nov 20 '14 at 5:33
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    $\begingroup$ Agreed - this is one of the finest 'simple' answers I've ever seen on this site. $\endgroup$ – Steven Stadnicki Nov 20 '14 at 17:55
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Using the sum of a finite geometric series twice:

$$5+55+555+\ldots+\overbrace{55...5}^{n\;\text{times}}=\sum_{k=0}^n\left(5\cdot 10^k+5\cdot10^{k-1}+\ldots+5\cdot 10+5\right)=$$

$$=5\sum_{k=0}^n\frac{10^{k+1}-1}9=...\text{etc.}$$

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    $\begingroup$ Shouldn't the summation be for k=0 up to n-1? $\endgroup$ – ypercubeᵀᴹ Nov 20 '14 at 14:31
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Here is a recurrence relation:

  • $a_0=5$
  • $a_n=10a_{n-1}+5(n+1)$

Converting this to a direct formula, you get: $$\frac{5\cdot10^{n+2}-45n-95}{81}$$

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    $\begingroup$ how do you convert to a direct formula? $\endgroup$ – Yaitzme Nov 18 '14 at 11:19
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    $\begingroup$ @Yaitzme: Open wikihow.com/Solve-Recurrence-Relations and go to Method 3 of 5: Polynomial (will be easier than explaining it in a comment). $\endgroup$ – barak manos Nov 18 '14 at 11:28
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    $\begingroup$ That's awesome. Exactly what I wanted ! $\endgroup$ – Yaitzme Nov 18 '14 at 11:32
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Hint: Rather than gathering the terms in $10^r$ together as you have, try first summing $10^{n-1}+10^{n-2}+\dots +1$ as a geometric progression - which should give you a term in $10^n$ plus a constant. The constants are easy to add, and the terms in $10^n$ are another geometric progression.

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I think something like

$$5\sum_{i=0}^n (n+1-i)10^{i}$$

should work.

Some explanation as how it works: First of all I rewrite as: $5(1 + 11 + 111 + \ldots)$. Then notice that I can construct this sum also with adding numbers of the form $10^i$, while considering that at each digit the number should get added multiple times depending on the length of the number. For example:

  • $n=0$: $\,5 \cdot 10^0 = 5$
  • $n=1$: $\,5 \cdot (2 \cdot 10^0 + 1 \cdot 10^1)= 5 \cdot (2 + 10) = 60$
  • $n=2$: $\,5 \cdot (3 \cdot 10^0 + 2 \cdot 10^1 + 1 \cdot 10^2)= 5 \cdot (3 + 20 + 100) = 615$

and so on...

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  • $\begingroup$ Shouldn't the 5 be inside the summation? $\endgroup$ – JLee Nov 16 '14 at 12:43
  • $\begingroup$ @JLee should not make any difference if it is inside or outside the summation $\endgroup$ – Loreno Heer Nov 16 '14 at 12:45
  • $\begingroup$ oh yeah. it wouldn't matter, but when n=1, when I plug in i=1, I get 50, and I should get 55, right? $\endgroup$ – JLee Nov 16 '14 at 12:49
  • $\begingroup$ @JLee in case of $n=1$. It calculates $5(2+1*10)=60$ which is correct. $\endgroup$ – Loreno Heer Nov 16 '14 at 12:52
  • $\begingroup$ When n=0, i goes from 0 to 0, and I get 5, which is correct. When n=1, i goes from 0 to 1, so I add that 5 to the answer I get when i=1. 5(1 + 1 - 1)10^1 = 50 (What am I doing wrong here?) $\endgroup$ – JLee Nov 16 '14 at 13:01
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$5+55+555+5555+55555+\cdots n$ terms, Sum of finite seqence of this pattern can be obtained by re-arranging this into geometric sequence.

$5$ should be taken common first, we get $$5× (1+11+111+1111+11111+\cdots n)$$

$1, 11,111,\ldots$ are formed as quotients of chains of 9 and 9 like below

$$5× (9/9+99/9+999/9+9999/9+\cdots n)$$ then 9 is taken common out.

$$5/9×(9+99+999+9999+\cdots n)$$

$1,11,111,1111,...$are converted into constitution of geometric sequence as $$(10-1), (100-1), (1000-1)\ldots$$then

$5/9×(10-1+100-1+1000-1+10000-1+\cdots n)$ then its divided into two groups as below

$$5/9×(10+100+1000+10000+\cdots n -1-1-1-1-1-\cdots n)$$

first one is geometric sequence with common ratio 10 and first term akso 10 then using sum of finite GS for n terms $S_n = a(r*n - 1)/(r-1)$ for $r\neq1$ and for series $-1-1-1-1-1\cdots -n$ terms gets (-n) then using these we get,

$$5/9*(10*(10^n-1)/(10-1)-n)$$

$$5/9×(10/9×(10^n-1)-n)$$

$$50/81×(10^n-1)-5n/9$$

as required answer.. doing similarily we ca get sum of sequences like below

$0.4 + 0.44 + 0.444 + 0.4444 + \cdots$ upto n terms.

thank you ..

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    $\begingroup$ Wow, that was increasibly unreadable. Check out this page for tips on how to write a post and this page for tips on how to write math formulas. $\endgroup$ – Winther Nov 17 '14 at 5:27
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    $\begingroup$ To start a new line, press Enter twice. $\endgroup$ – Akiva Weinberger Nov 17 '14 at 5:44
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    $\begingroup$ @Wither Thank you so much for redirecting me by giving such a good suggestion..I am going through that page its really great.. $\endgroup$ – Mangal Bajracharya Nov 17 '14 at 5:54
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    $\begingroup$ @columbus8myhw thank u.. for this suggestion next time i will definitely gonna use it.. $\endgroup$ – Mangal Bajracharya Nov 17 '14 at 5:56
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    $\begingroup$ "next time i will definitely gonna use it" Great, why not start with this one (still unreadable)? $\endgroup$ – Did Nov 17 '14 at 8:12
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$$5=5*1$$ $$5+55=5*12$$ $$5+55+555=5*123$$ $$5+55+555+5555=5*1234$$ $$5+55+555+5555+55555=5*12345$$ $$5+55+555+5555+55555+555555=5*123456$$ $$...............$$ $$...............$$ $$5+55+555+5555+55555+555555+5555555+55555555+555555555=5*123456789$$

enter image description here

$$5+55+555+5555+55555+555555+5555555+55555555+555555555+5555555555=5*1234567900$$ enter image description here and so on

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This might work:

It's adding together powers of 10 multiplied by 5, and then adding together those numbers.

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    $\begingroup$ This isn't really a formula; it's more of a formal restatement of the problem. $\endgroup$ – Ypnypn Nov 18 '14 at 2:32
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    $\begingroup$ @Ypnypn For a beginner, restating the problem algebraically is an important step. Well, that's true for everyone actually. $\endgroup$ – DanielV Nov 18 '14 at 23:00
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    $\begingroup$ This works perfectly. The inner sum is a geometric progression and yields a power of $10$ and a constant. So the outer sum is another geometric progression a constant progression, yielding a power of $10$, a constant and a linear term. $\endgroup$ – Yves Daoust Nov 19 '14 at 13:18
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    $\begingroup$ I like this. It's exactly what first came to mind when I read the question. $\endgroup$ – ventsyv Nov 19 '14 at 20:20
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Obviously for the general term of the sum, $t_n=10t_{n-1}+5$, with $t_0=5$.

By identifying with $t_n+c=10(t_{n-1}+c)+5$, you can rewrite this recurrence as $$t_n+\frac59=10(t_{n-1}+\frac59),$$ then $$t_n+\frac59=10(t_{n-1}+\frac59)=100(t_{n-2}+\frac59)...=10^{n}(t_0+\frac59)=10^n\frac{50}9.$$ Now you can sum these terms, using the formula for the geometric series:

$$\sum_{i=0}^{n-1} t_i=\sum_{i=0}^{n-1}\left(10^{n}\frac{50}9-\frac59\right)=\frac{10^n-1}{9}\frac{50}9-n\frac59.$$

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$$\begin{align}5+55+555+5555+...&=5(1+11+111+1111+...)\\&=\frac 59(9+99+999+9999+...)\\&=\frac 59\sum_{i=1}^n 10^i-1\\&=\frac 59(\sum_{i=1}^n 10^i-\sum_{i=1}^n 1)\\&=\frac 59((\sum_{i=0}^n 10^i)-1-\sum_{i=1}^n 1)\\&=\frac 59\left(\frac {10^{n+1}-1}{10-1}-1-n\right)\\&=\frac 59\left(\frac {10^{n+1}-9n-10}9\right)\\&=\frac{50(10^{n}-1)-45n}{81}\end{align}$$

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$$\begin{align} S&=\;\;\;\underbrace{5+55+555+\cdots+\overbrace{555...55}^{n}}_{\text{$n$ terms}}\\ 10S&=\quad\;\;\;\; \ 50+550+\cdots +555...50+\overbrace{5555...5}^{n}0\\ \\ 10S-S&=-5\;\;-5\quad -5\quad\cdots\qquad\;-5+\overbrace{5555...5}^{n}0\\ \\ 9S&=-5n+\overbrace{5555...5}^{n}0\\ \\ S&=\frac59 (\overbrace{1111...1}^{n}0-n)\qquad\blacksquare \end{align}$$

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$(\frac{5}{9})(10^n) - \frac{5}{9}$

$\frac{5}{9}$ is .555 repeating. $10^n$ will shift the desired number of 5's past the decimal place. Then subtracting $\frac{5}{9}$ will clip the remaining 5 repeating past the decimal place.

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The first order difference of the series is $\Delta_n^1=55,555,5555,55555...$.

The second order difference is $\Delta_n^2=500,5000,50000...=50.10^n$.

As the finite difference of a power is the same power, we have (adding an affine term that vanishes with the second order difference)

$$S_n=a.10^n+bn+c,$$ $$\Delta_n^1=a.9.10^n+b,$$ $$\Delta_n^2=a.81.10^n.$$

Compute the unknowns from $\Delta_1^2=500, \Delta_1^1=55, S_1=5$ to get $$S_n=\frac{50}{81}.10^n-\frac59n-\frac{50}{81}.$$

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$$S = 5+55+555+5555+...$$ $$S = 5(1+11+111+1111+...)$$ $$S = 5\left(\sum_{k=0}^010^k+\sum_{k=0}^110^k+\sum_{k=0}^210^k+\sum_{k=0}^310^k+\dots \sum_{k=0}^n10^k\right)$$ Let $$\sum_{k=0}^n10^k = \frac{1}{9}(10^{n+1}-1)$$ Which implies \begin{align} S &= 5\sum_{k=0}^n\left[ \frac{1}{9}(10^{k+1}-1)\right]\\ &= \frac{5}{9}\sum_{k=0}^n(10^{k+1}-1)\\ &= \frac{5}{9}\left(\sum_{k=0}^n10^{k+1}-\sum_{k=0}^n1\right)\\ &= \frac{5}{9}\left(10\sum_{k=0}^{n}10^{k}-(n+1)\right)\\ &= \frac{5}{9}\left(\frac{10}{9}(10^{n+1}-1)-n-1)\right) \end{align} Therefore

$$S = \frac{5}{81}\left(10^{n+2}-9n-19)\right)$$

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