Basic summation issue

Asked Jun 23 '23 at 21:42

Active Jun 23 '23 at 21:42

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I am having some trouble prooving the following summation formula. As a and b are constants I got the constant part out of the sum and tried to split the summation (such that sum of a to b is the same as sum of 1 to b minus sum of 1 to a plus a for example), but it always ends up with an expression that never simplifies to this result. I also spent a lot of time looking for the formula for this summation and somehow cannot manage to find exactly this:

$$\sum_{i=a}^b i \cdot \frac{1}{b-a+1} = \frac{a+b}{2}$$

How do we get from left to right here? Or even simpler, what is the formula for just this sum?

$$\sum_{i=a}^b i$$

asked Jun 23 '23 at 21:42

evilmandarine

Hint: The sum is an arithmetic series with common difference 1. How many terms does the series have? – Yaneda Jun 23 '23 at 21:45
It has (b-a+1) right? – evilmandarine Jun 23 '23 at 21:49
yup! So what does the summation have to equal? – Yaneda Jun 23 '23 at 21:50
1

Use Gauss's trick : If we write down the sum twice , once from $a$ to $b$ , then from $b$ to $a$ we get $b-a+1$ pairs summing up to $a+b$. Hence the value of the sum is $\frac{(a+b)(b-a+1)}{2}$ – Peter Jun 23 '23 at 21:52
((b-a+1)(1 + (b-a+1))) / 2. I see the idea. But at the end I am at (b-a+2)/2 :( I'm gonna cry soon. – evilmandarine Jun 23 '23 at 21:53
Peter gave you the answer :) – julio_es_sui_glace Jun 23 '23 at 21:58
1

@Peter ok I put that on paper and finally I got it, thank you so much. I was probably over-complicating it trying sum(1 to b) minus sum(1 to a) plus 1 using the n(n+1)/2 formula. I don't get why that did not work though. – evilmandarine Jun 23 '23 at 22:04
@AnneBauval it does exactly, so this question is kind of a duplicate, sorry. But there is no way I know how to even begin to look for this, can you look for MathJax? – evilmandarine Jun 23 '23 at 22:09
1

For your approach,$$\begin{align} \sum_{i=1}^bi - \sum_{i=1}^ai + a &= \frac{b(b+1)}{2} - \frac{a(a+1)}{2} +a\ &= \frac{b(b+1) - a(a+1) + 2a}{2}\ &= \frac{b^2-a^2 + b+a}{2}\ &= \frac{(b-a)(b+a) + b+a}{2}\ &= \frac{(b-a+1)(a+b)}{2}\ \end{align}$$ – peterwhy Jun 23 '23 at 22:09
The duplicate was found using Approach$0$. – Anne Bauval Jun 23 '23 at 22:12
@peterwhy Ok I now see what I was missing thanks to your comment... +1 – evilmandarine Jun 23 '23 at 22:14

Basic summation issue

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