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# Series: $$a+b+a+b...+a+b, b>=a$$

Consider the series $$a+b+a+b+...+a+b+a+b, b >=a$$

A general expression for the nth sum of this series is given by:

$$n (\frac{a+b}{2}) - (\frac{b-a}{2}(n(mod 2))$$)

When $$a=b$$ this generalizes to $$a \times n$$

Can anyone point me to more information about this series? It may be interesting to find an expression for the $$nth$$ term of $$a+b+c+....+a+b+c, a<=b<=c$$ and so on for more terms $$(a,b,c,...,x,y,z)$$ etc.

I came up with this expression myself and realize I might be over complicating something simple. Any information is appreciated, especially with respect to notation.

Note by Steven Kidd
3 years, 6 months ago

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With notation I personally dont really like your use of mods although I understand what you mean. I would write: $$U_{n} = n(a+b)/2 + (a-b)/4 + ((a-b)/4)(-1)^{n}$$. Now for the generalisation: There are many better mathematicians on this site than me but I will give you my solution, leaving the proof to you (unless you ask me to explain): So, let k be the number of distinct terms (ie. a+b+c+a+b+c.... would be k= 3), n be the number of terms and we also have i where $$i \equiv n \pmod{k}$$, and also $$i \ne {0}$$. In this case we will call the distinct numbers $$a_{1}$$, $$a_{2}$$, $$a_{3}$$ etc. instead of a, b, c My formula would be:( $$a_{1} \times (2n-i)$$ + $$a_{2} \times (2n-i)$$ + $$a_{3} \times (2n-i)$$ + ... + $$a_{i-1} \times (2n-i)$$ + $$a_{i} \times (2n-i)$$ + $$a_{i+1} \times (n-i)$$ + $$a_{i+2} \times (n-i)$$ + $$a_{i+3} \times (n-i)$$ + ... + $$a_{k} \times (n-i)$$ ) / k. If $$i = 0$$ then it is trivial. · 3 years, 6 months ago

I've been looking back over this and realised I made a few errors. First of all, in the case where $$k = 2$$ it should be $$-1^{n+1}$$ instead of $$-1^{n}$$. As for the generalisation, I made a fairly stupid misrake, it should be: ( $$a_{1} \times (n+k-i)$$ + $$a_{2} \times (n+k-i)$$ + ... + $$a_{i} \times (n+k-i)$$ + $$a_{i+1} \times (n-i)$$ + $$a_{i+2} \times (n-i)$$ + ... + $$a_{k} \times (n-i)$$ )/ k · 3 years, 6 months ago

I like how you got rid of the mod, and it makes me wonder weather the same sort of thing could be done to get rid of a $$- (n(mod k))$$ term as i had used it in a general sense, (n,k positive integers, n <= k). I'll post back once I get some real time to go over this if I have any further questions, thanks :) · 3 years, 6 months ago

I have no idea how to do it efficiently, but I came up with this off the top of my head:

$n\text{ (mod }k)=\dfrac{k\cdot \text{arg}\left(e^{2\pi i n/k}\right)}{2\pi}$

Feel free to simplify it if you can find a way. · 3 years ago