Series: a+b+a+b...+a+b,b>=aa+b+a+b...+a+b, b>=a

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

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

n(a+b2)(ba2(n(mod2))n (\frac{a+b}{2}) - (\frac{b-a}{2}(n(mod 2)))

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

Can anyone point me to more information about this series? It may be interesting to find an expression for the nthnth term of a+b+c+....+a+b+c,a<=b<=ca+b+c+....+a+b+c, a<=b<=c and so on for more terms (a,b,c,...,x,y,z)(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
6 years, 5 months ago

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

steven kidd - 6 years, 5 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: Un=n(a+b)/2+(ab)/4+((ab)/4)(1)n 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 in(modk) i \equiv n \pmod{k} , and also i0 i \ne {0} . In this case we will call the distinct numbers a1 a_{1} , a2 a_{2} , a3 a_{3} etc. instead of a, b, c My formula would be:( a1×(2ni) a_{1} \times (2n-i) + a2×(2ni) a_{2} \times (2n-i) + a3×(2ni) a_{3} \times (2n-i) + ... + ai1×(2ni) a_{i-1} \times (2n-i) + ai×(2ni) a_{i} \times (2n-i) + ai+1×(ni) a_{i+1} \times (n-i) + ai+2×(ni) a_{i+2} \times (n-i) + ai+3×(ni) a_{i+3} \times (n-i) + ... + ak×(ni) a_{k} \times (n-i) ) / k. If i=0 i = 0 then it is trivial.

Josh Rowley - 6 years, 4 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 k = 2 it should be 1n+1 -1^{n+1} instead of 1n -1^{n} . As for the generalisation, I made a fairly stupid misrake, it should be: ( a1×(n+ki) a_{1} \times (n+k-i) + a2×(n+ki) a_{2} \times (n+k-i) + ... + ai×(n+ki) a_{i} \times (n+k-i) + ai+1×(ni) a_{i+1} \times (n-i) + ai+2×(ni) a_{i+2} \times (n-i) + ... + ak×(ni) a_{k} \times (n-i) )/ k

Josh Rowley - 6 years, 4 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(modk)) - (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 - 6 years, 4 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 (mod k)=karg(e2πin/k)2πn\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 - 5 years, 10 months ago

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