If the sum of all positive even integers less than $1000$ is $ A $ , what is the sum of all positive odd integers less than $1000$?
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$\begingroup$ You can first use the formula for the sum of the first $ \ n \ $ integers to find the sum of all the positive integers to 1000. Then use the same formula for the sum of all the positive integers to 500 and double that; that is the sum of all the even integers to 1000. The difference is the sum you are interested in. [OK, looks like I have now replied to the original version of this question...] $\endgroup$– colormegoneCommented Jun 12, 2013 at 18:58
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11$\begingroup$ The sum of all positive odd integers below 1000 is what it is, whether or not the sum of the even ones is denoted by the symbol A. The question needs to say "... what is the sum of all positive odd integers in terms of A?". $\endgroup$– KazCommented Jun 12, 2013 at 20:14
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3$\begingroup$ If the sum of all positive even integers less than $1000$ is $A$, then I'd say the sum of all positive odd integers less than 1000 is $B$. ;-) $\endgroup$– celtschkCommented Jun 12, 2013 at 20:56
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$\begingroup$ Heh, first thing in the morning… I read "Why is the sum of all positive integers less than 1000?" and expected inductive proof :P $\endgroup$– PotatoswatterCommented Jun 12, 2013 at 22:00
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$\begingroup$ @Kaz: That's still not specific enough. It seems that the intended answer is $A + 500$, but $\frac{500A}{499}$ would be equally valid. For that matter, so would ceil$(A^{1.000161})$. $\endgroup$– DanCommented Jun 13, 2013 at 2:04
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9 Answers
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\begin{align} \underbrace{1}_{0+1}+\underbrace{3}_{2+1}+5+\cdots+\underbrace{999}_{998+1}&=\underbrace{(0+1)+(2+1)+\cdots+(998+1)}_{\text{500 terms}}=\\ &=\underbrace{0+2+4+\cdots+998}_{A}+\underbrace{1+1+\cdots+1}_{500 \text{ ones}}=A+500 \end{align}
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$\begingroup$ Hmm something is wrong , it says A+500 $\endgroup$– SSKCommented Jun 12, 2013 at 19:01
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1$\begingroup$ Ron Ford, are you another misspelling of Rob Ford? $\endgroup$ Commented Apr 28, 2014 at 15:13
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If the sum of all even numbers is $A$ then $2+4+\ldots+998=A$. Subtracting $1$ from each of these gives $1+3+\ldots+997=A-499$ and adding $999$ to that gives $A+500$.
answered Jun 12, 2013 at 19:04
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$\begingroup$ @SSK We have to add $999$ to $A-499$ because $A-499$ is the sum $1+3+...+997$. Since $999$ is also an odd number less than $1000$, we need it. $\endgroup$ Commented Jun 12, 2013 at 19:46
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3$\begingroup$ Why do two steps? Why not directly note that: $$(0 + 2 + \dots + 998) + 500 = (0 + 1) + (2 + 1) + \dots + (998 + 1) = 1 + 3 + 5 + \dots + 999$$ $\endgroup$– TMMCommented Jun 12, 2013 at 20:20
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$$2+4+6+\ldots+994+996+998=A\ \ \ \ \ /-499$$ $$(2-1)+(4-1)+(6-1)+\ldots+(994-1)+(996-1)+(998-1)=A-499 /\mbox{substracting in brackets}$$ $$1+3+5+\ldots+993+995+997=A-499\ \ \ \ \ /+999$$ $$1+3+5+\ldots+993+995+997+999=A+500.$$ Thus if sum of all positive even integers less than $1000$ is $A$, then sum of all odd integers less then $1000$ is $A+500$.
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4$\begingroup$ if you start the even series at
0and go to998, you can then add1to every value to get the odd series, it's also trivial to show that that series contains500values. $\endgroup$– zzzzBovCommented Jun 12, 2013 at 20:38
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You look for this sum:
$$\sum_{k=0}^{499}{(2k+1)}=2\sum_{k=0}^{499}k+499=2\frac{499\times 500}{2}+499=500^2$$
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$$\begin{align*}\color{green}{1}+\color{blue}{3}+\color{red}{5}+\ldots+\color{red}{995}+\color{blue}{997}+\color{green}{999}&=\color{green}{(1+999)}+\color{blue}{(3+997)}+\color{red}{(5+995)}+\ldots+\color{magenta}{(499+501)}=\\&=250\cdot{1\,000}={250\,000}\color{grey}{=500^2}\end{align*}$$
In the penultimate step, we used that there are $250$ positive odd numbers less than $500$.
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These equations will give you the sum of all numbers 1 Through $n-1$, since your requirement is "Under 1000". The special thing with my answer(s) is that they will work with both Odd AND Even numbers. So if you input 967, the equations would indeed work from 1 to 966. And for your requirement... if you put in 1000, you get answers for 1 to 999
- Assuming $n$ = Any Whole Number (that will not be included)
- Assuming $A$ = (Sum of Even Numbers < $n$) - $\left(\lfloor{\left( n - 1 \right) \over 2}\rfloor + 1\right)*\lfloor{(n-1)/2}\rfloor$
- Assuming $B$ = (Sum of Odd Numbers < $n$), as it relates to $A$
- Solution: $B = A + \left( \lceil { n - 1 \over 2 } \rceil * \left( 1-2*MOD(n,2) \right) \right)$
- Also, Assuming $C$ = (Sum of Even Numbers < $n$), as it relates to $B$
- Inverse Solution: $A = B - \left( \lceil { n - 1 \over 2 } \rceil * \left( 1-2*MOD(n,2) \right) \right)$
Also, the sum of All Numbers Under $n$ is:
$${n*(n+1) \over 2} - n$$
The sum of all Even Numbers Under $n$ is:
$$\left(\lfloor{\left( n - 1 \right) \over 2}\rfloor + 1\right)*\lfloor{(n-1)/2}\rfloor$$
The sum of all Odd Numbers Under $n$ is:
$$\left( \lceil{n-1 \over 2}\rceil - 1 \right) ^ 2 + \left( n-\left(1+MOD(n,2)\right) \right)$$
Discuss. ^_^
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$\begingroup$ I think many of your formulae are wrong. The formula for $B$ looks wrong and the final formula does as well. $\endgroup$– RotsorCommented Jun 12, 2013 at 22:29
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$\begingroup$ Thanks for using the formatting. I looked at the source in "edit" and I think I understand how to format mathmatical notation now. I appreciate it! Anyway, I did a lot of testing of these equations and they appear to be correct as far as I can tell. $\endgroup$– SuamereCommented Jun 13, 2013 at 14:25
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$\begingroup$ Well, let's assume $n = 1000$. Then $A$, the sum of even numbers, is $250500$ and $B$, the sum of odd numbers, is $250000$. Then you are saying $B = A + \lceil{A \over 2 } \rceil$, but the right-hand side equals to $375750$, which is different from left-hand side. Also your final formula evaluates to $125250^2 = 15687562500$, which is clearly wrong. $\endgroup$– RotsorCommented Jun 13, 2013 at 14:29
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$\begingroup$ You were correct for the most part. In my tests I was putting in 999. So for most of the equations, I changed n to n-1 so the user can put in 1000. Also, I put A in the last equation instead of n. So in short, I made some changes to fix my errors which mostly included assumptions (which are bad). I tested the equations in Excel and on Wolfram. With 1000 as the input:
Total=499500/Evens = 249500/Odds = 250000/Odds (In relation to Evens) also = 250000Thanks for keepin' me honest. ^_^ $\endgroup$– SuamereCommented Jun 13, 2013 at 15:17 -
$\begingroup$ lol, by replacing $n$ with $\left(\lceil{n \over 2}\rceil*2\right)$, it wouldn't matter if you put in 999 or 1000 for $n$. But that's overkill I think... $\endgroup$– SuamereCommented Jun 13, 2013 at 15:30
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The first $1000$ odd integers run from $1$ to $999$.
If we add up the extreme integers we get:
$$1 + 999 = 1000$$ $$3 + 997 = 1000$$ $$\vdots$$
we notice that the all add up to $1000$. We also know that there are $\frac{1000}{2} = 500$ pairs of numbers and so the number of pairs of odd numbers is going to be half this, i.e $250$ and so the sum of the integers is going to be
$$1000 \times 250 = 250,000.$$
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If a is how you said, then the answer would be $A + 500$ as $1 + 3+\ldots + 997$ is $A - 499$, plus $999$ and you get $A + 500$.
Good luck on your homework.
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Consider the integers from $1$ to $10$. The even integers$\,\,(2,4,6,8)$ sum is $20$ and the odd integers $(1,3,5,7,9)$ is $25$. The difference is $5$. For the sum of the odd integers below $1000 =10 \times 100$, we need to add $\,500=5\times 100$. Therefore, the answer is $A+500$.
