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First of all: beginner here, sorry if this is trivial.

We know that $ 1+2+3+4+\ldots+n = \dfrac{n\times(n+1)}2 $ .

My question is: what if instead of moving by 1, we moved by an arbitrary number, say 3 or 11? $ 11+22+33+44+\ldots+11n = $ ? The way I've understood the usual formula is that the first number plus the last equals the second number plus second to last, and so on. In this case, this is also true but I can't seem to find a way to generalize it.

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  • $\begingroup$ Do you mean $ 11+22+33+44+\ldots+\color{red}{11}n $ ? $\endgroup$ – callculus Apr 18 '16 at 8:13
  • $\begingroup$ @calculus Yes, you're right, sorry about that. $\endgroup$ – ChatterOne Apr 18 '16 at 8:16
  • $\begingroup$ No problem. It was just for clarification. I´m just wondering how your question could be answered 5 times with such a typo. I fixed it. $\endgroup$ – callculus Apr 18 '16 at 8:20
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    $\begingroup$ Try multiplying your first equation by 11 on both sides $\endgroup$ – Abraham Zhang Apr 18 '16 at 8:23
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This a question of notation.

$1+2+3+4+\dots+n$ is a notation for $\sum_{k=1}^n k$

I assume that $11+22+33+44+\dots+11\times n$ is a notation for $\sum_{k=1}^n 11\times k$

in this case, you just get : $$ \sum_{k=1}^n 11\times k = 11 \times \sum_{k=1}^n k = 11 \frac{n(n+1)}{2}$$

with any $M=3$ or $11$, you get $\sum_{k=1}^n M\times k= M \times \sum_{k=1}^n k$

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  • $\begingroup$ Being a beginner, this is the easiest answer for me to understand, even though all the other answers are valid of course. $\endgroup$ – ChatterOne Apr 18 '16 at 8:24
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    $\begingroup$ This is not a question of notation. $11+22+33+44+\dots+n\color{red}{\neq }\sum_{k=1}^n 11\times k$ $\endgroup$ – callculus Apr 18 '16 at 8:27
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    $\begingroup$ $11+22+33+44+\dots+n$ means nothing in itself $\endgroup$ – MJ73550 Apr 18 '16 at 8:39
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The general formula derives from the simple one.

The general sequence is $$a=a-b+b,a+b=a-b+2b,a+2b,\cdots a-b+nb,$$ i.e. $n$ terms from $a$ to $a+(n-1)b=a+c$.

Then

$$\sum_{k=1}^n(a-b+kb)=n(a-b)+\frac{n(n+1)}2b=n\frac{a+a+(n-1)b}2=n\frac{a+c}2.$$


With $11$ to $11n$ in steps $11$,

$$n\frac{11+11n}2=11\frac{n(1+n)}2.$$

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You get what is called an arithmetic series. More details are at https://en.wikipedia.org/wiki/Arithmetic_progression .

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Assume you have a sequence $$x, x+a, x+2a, x+3a, ..., x+na.$$

Let us note $$S = \sum_{i=0}^n (x+ia).$$ Then by the trick you mentioned, we see that $$S+S = (2x+na)+(2x+na)+...+(2x+na) = (n+1)(2x+na).$$ Hence $$S = \frac{(n+1)(2x+na)}{2}.$$

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The key here is to look at the constant(!) differences. In $1+2+3+4+...n$ it is $d=1$, and the last number is, calling the first number $a$, $a+(n-1)*d=1+n-1=n$. In your next example the difference is $d=11$, and with $a=11$ you have the last number $a+(n-1)*d=11+(n-1)*11 = 11n $ and the sum can then analogically be computed like the first sum.

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There's a simple rule for linear progressions: The sum equals the number of items, times the average of the first and last item. So in this case n (11n + 11) / 2.

If you added 7, 17, 27, 37 and so on, the first item would be 7, the last would be 10n - 3, the average (10n - 3 + 7) / 2 = 5n + 2, so the sum would be n (5n + 2).

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