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
5 years, 10 months ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

Sort by:

Top Newest

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

steven kidd - 5 years, 10 months ago

Log in to reply

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 - 5 years, 10 months ago

Log in to reply

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 - 5 years, 10 months ago

Log in to reply

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 - 5 years, 10 months ago

Log in to reply

@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, 4 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...