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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, 9 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.

Josh Rowley - 3 years, 9 months ago

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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

Josh Rowley - 3 years, 9 months ago

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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 :)

Steven Kidd - 3 years, 9 months ago

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@Steven Kidd 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.

Daniel Liu - 3 years, 3 months ago

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a,b,c... etc are positive integers

Steven Kidd - 3 years, 9 months ago

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