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I recently came along a question regarding a series where it was required to write a general formula to calculate the sum when the number of terms is given. The series was as follows - 1+11+111+1111... and so on.

I immediately noticed that the sum of 1+11+111+1111... could be expressed as an AGP [$10^{n-1}$+2($10^{n-2}$)+3($10^{n-3}$)+...n] with the first term being $10^{(n-1)}$, the common difference as 1 and the common ratio as 0.1 with n denoting the number of terms.

Inputting these values into the formula for the sum of an AGP uptil n terms, I got - $$\frac{10^n}{9} + \frac{10^n - 10^{2-n}}{81} - \frac{[1+(n-1)10^{1-n}]}{9}$$ as the resultant formula.

Formula used as reference to calculate the sum of an AGP - $$\frac{a}{1-r} + \frac{dr(1-r^{n-1})}{(1-r)^2}-\frac{[a+(n-1)d]r^n}{1-r}$$ where a denotes the initial term of the AGP, r denotes the common ratio, and d denotes the common difference.

However, this formula only works for n=1, and for all succeeding integral values of n, there always exists some kind of error. For example, at n=2, the formula outputs the value of 11.1, at n=3, it outputs 111.12.

I've gone through the formula multiple times, and am unable to point out the error. Hence, I would be highly obliged if any of you could tell me whether I have committed a mistake, or if this is a problem with the formula of a sum of AGP uptil n terms.

Thanks, a math nerd.

P.S. I apologize for any and all formatting mistakes, as I am very new to this MathJax stuff.

Note - I have managed to figure out my blunder (I had taken a to be equal to $10^{n-1}$ and so the resultant series wasn't an AGP) and so this question no longer requires an answer. Thanks to any & all who took out their precious time to look at the problem.

Divyansh Bajpai
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  • Well, here are the possibilities: The formula is wrong. You have copied the formula incorrectly. You have misunderstood the formula. You have made an arithmetical error. Mathematics is inconsistent, and we can all go do something else instead. Now, unless you tell us what formula you are using, and what the letters in it stand for, it's hard for us to distinguish among these possibilities (although I'm pretty much ready to rule out the collapse of mathematics). Also, there is help with formatting available on this site. – Gerry Myerson May 30 '24 at 06:30
  • Help with formatting: https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – Gerry Myerson May 30 '24 at 06:30
  • here is an earlier AGP question, with links to even earlier AGP questions. Maybe there will be something useful at one of them. https://math.stackexchange.com/questions/2441307/hints-on-solving-sum-i-1n-i-cdot-2i-1 also https://math.stackexchange.com/questions/3828189/summation-problem-s-n-sum-limits-k-2n6k-2-kn2 and https://math.stackexchange.com/questions/1994459/how-do-we-find-the-sum-of-this-series-sum-an-1d-b-rn-1 – Gerry Myerson May 30 '24 at 06:35
  • And https://math.stackexchange.com/questions/1955682/how-is-this-sequence-in-arithmetico-geometric-progression-agp – Gerry Myerson May 30 '24 at 06:41
  • @GerryMyerson Apologies for the earlier confusion. I've gone back and edited my original question to include the formula which I had used as a reference to calculate the sum of the AGP. – Divyansh Bajpai May 30 '24 at 07:46
  • OK, so all that's missing now is, how do you define the common ratio and the common difference of an AGP? – Gerry Myerson May 30 '24 at 12:25
  • Put it another way, when $n=2$, the formula you have given says the sum is $a-(a+d)r$. Is that what a two-term AGP with common ratio $r$ and common difference $d$ looks like? If not, then you have the wrong formula. Have you looked at any of the links I gave, to see whether they give a formula for the sum of an AGP? – Gerry Myerson May 30 '24 at 12:32
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    I have just gone back and noticed a huge blunder on my part. I had taken a to be equal to 10^(n-1), and so the resultant series is in actuality, not an AGP. In fact, to convert it into one, you'd have to take 10^(n-1) common from all the terms leaving behind the AGP [(1+2/10+3/100...n/(10^n-1)]. When we use the formula to compute this sum, we get the right answer upon multiplication by 10^(n-1). – Divyansh Bajpai May 30 '24 at 16:21

1 Answers1

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$$S = 1+11+111+1111+\cdots+11\cdots11$$ $$S = \frac{1}{9}\left(9+99+999+9999+\cdots+99\cdots99\right)$$ $$S = \frac{1}{9}\left(10-1+100-1+1000-1+10000-1+\cdots+1000\cdots-1\right)$$ $$S = \frac{1}{9}\left(10-1+100-1+1000-1+10000-1+\cdots+1000\cdots-1\right)$$ $$S = \frac{1}{9}\left(10+100+1000+10000+\cdots+1000\cdots-n\right)$$ $$S = \frac{1}{9}\left(\frac{10(10^n-1)}{9}-n\right)$$ $$S_n =\frac{10(10^n-1)-9n}{81}$$

$$S_1 =\frac{10(10^n-1)-9n}{81}=\frac{10(10-1)-9}{81}=1$$ $$S_2 =\frac{10(10^n-1)-9n}{81}=\frac{10(100-1)-18}{81}=12$$ $$S_3 =\frac{10(10^n-1)-9n}{81}=\frac{10(1000-1)-27}{81}=123$$ $$S_4 =\frac{10(10^n-1)-9n}{81}=\frac{10(10000-1)-36}{81}=1234$$

Amrut Ayan
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    I have already reached this solution using the formulas pertaining to GP. My original question was moreso asking why we can't reach the same via converting the given GP into an AGP. However, I appreciate your answer nonetheless. Thank you. – Divyansh Bajpai May 30 '24 at 08:32