What is sum of all positive odd integers less than $1000$? If the sum of all positive even integers less than $1000$ is $ A $ , what is the sum of all positive odd integers less than $1000$? 
 A: You look for this sum:
$$\sum_{k=0}^{499}{(2k+1)}=2\sum_{k=0}^{499}k+499=2\frac{499\times 500}{2}+499=500^2$$
A: $$\begin{align*}\color{green}{1}+\color{blue}{3}+\color{red}{5}+\ldots+\color{red}{995}+\color{blue}{997}+\color{green}{999}&=\color{green}{(1+999)}+\color{blue}{(3+997)}+\color{red}{(5+995)}+\ldots+\color{magenta}{(499+501)}=\\&=250\cdot{1\,000}={250\,000}\color{grey}{=500^2}\end{align*}$$
In the penultimate step, we used that there are $250$ positive odd numbers less than $500$.
A: \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}
A: 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 $A+500$.
A: $$2+4+6+\ldots+994+996+998=A\ \ \ \ \ /-499$$
$$(2-1)+(4-1)+(6-1)+\ldots+(994-1)+(996-1)+(998-1)=A-499 /\mbox{substracting in brackets}$$
$$1+3+5+\ldots+993+995+997=A-499\ \ \ \ \ /+999$$
$$1+3+5+\ldots+993+995+997+999=A+500.$$
Thus if sum of all positive even integers less than $1000$ is $A$, then sum of all odd integers less then $1000$ is $A+500$.
A: 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.  ^_^
A: 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
$$1000 \times 250 = 250,000.$$
A: 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.
A: 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 $A+500$.
