# Odds and Evens

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• Mar 5th 2011, 01:46 AM
dumluck
Odds and Evens
Hi All,
I'm struggling understanding multiple points in the following question and was hoping for some help...

Q. For any positive integer n, the sum of the first n positive integers equals: $\frac{n(n+1)}{2}$ what is the sum of all the even integers between 99 and 301.

Ok. so the first bit is fine. 99-301 means we are looking at even integers 100-300.

q1: Looking at the formula am i right in saying that it denotes $\frac{integer(next integer)}{2} = SUM$ ?

q2: The next part I find uber confusing.

$2(\frac{150(150+1)}{2}) - 2(\frac{49(49+1)}{2})$

q2a. 150 is the 150th even number, why is that important?
q2b. I have no idea where the 49 came from and why it is important?
q2c: we are dividing by 2, I understand 2 is an even number but is this saying that if you divide ANY sum by 2 (or an even number) it will give you the sum of the even numbers in that sum?
q2d: why are we multiplying each result by 2?

It could just be that I'm not understanding the question, which is equally as important.

• Mar 5th 2011, 02:36 AM
emakarov
Quote:

q1: Looking at the formula am i right in saying that it denotes \frac{integer(nextinteger)}{2}=SUM?
Yes.

100 + 102 + ... + 300=2(50 + 51 + ... + 150) = 2((1 + 2 + ... + 150) - (1 + 2 + ... + 49)) = 2(150 * 151 / 2 - 49 * 50 / 2).
• Mar 5th 2011, 03:11 AM
Wilmer
Sum of first n even numbers = n(n + 1) ; example, 1st 4:
2 + 4 + 6 + 8 = 4(5) = 20
• Mar 5th 2011, 04:45 AM
Soroban
Hello, dumluck!

Quote:

$\text{For any positive integer }n,$
$\text{the sum of the first }n\text{ positive integers equals: }\:\dfrac{n(n+1)}{2}$

$\text{What is the sum of all even integers between 99 and 301?}$

Ok. so the first bit is fine. 99-301 means we are looking at even integers 100-300.

q1: Looking at the formula am i right in saying that it denotes:

/ . $\dfrac{\text{integer}(\text{next integer})}{2} \,=\, \text{Sum}\,?$ . Yes

Quote:

q2: The next part I find uber confusing: . $2\left(\dfrac{150(150+1)}{2}\right) - 2\left(\dfrac{49(49+1)}{2}\right)$

q2a. 150 is the 150th even number, why is that important?

q2b. I have no idea where the 49 came from and why it is important?

q2c: we are dividing by 2, I understand 2 is an even number
but is this saying that if you divide ANY sum by 2 (or an even number)
it will give you the sum of the even numbers in that sum?

q2d: why are we multiplying each result by 2?

I believe I can answer all your questions by deriving that formula.

$\text{We want the sum of even numbers from 100 to 300: }S$

$\text{First, find the sum of }all\text{ even numbers from 2 to 300: }S_1$

$\text{Note that: }\:S_1 \;=\; 2 + 4 + 6 + \hdots + 300 \;=\;2(1 + 2 + 3 + \hdots + 150)$

$\text{Hence: }\:S_1 \:=\:2(\text{sum of first 150 numbers}) \;=\;2\left(\dfrac{150(150+1)}{2}\right)$

$\text{We want to }subtract\text{ the sum of even numbers from 2 to 98: }S_2$

$\text{Note that: }\:S_2 \;=\;2 + 4 + 6 + \hdots + 98 \;=\;2(1 + 2 + 3 + \hdots + 49)$

$\text{Hence: }\:S_2 \:=\:2(\text{sum of first 49 numbers}) \;=\;2\left(\dfrac{49(49+1)}{2}\right)$

$\text{Therefore: }\:S \;=\;S_1 - S_2 \;=\;2\left(\dfrac{150(150+1)}{2}\right) - 2\left(\dfrac{49(49+1)}{2}\right)$

• Mar 5th 2011, 05:30 AM
dumluck
Quote:

Originally Posted by Soroban
Hello, dumluck!

I believe I can answer all your questions by deriving that formula.

$\text{We want the sum of even numbers from 100 to 300: }S$

$\text{First, find the sum of }all\text{ even numbers from 2 to 300: }S_1$

$\text{Note that: }\:S_1 \;=\; 2 + 4 + 6 + \hdots + 300 \;=\;2(1 + 2 + 3 + \hdots + 150)$

$\text{Hence: }\:S_1 \:=\:2(\text{sum of first 150 numbers}) \;=\;2\left(\dfrac{150(150+1)}{2}\right)$

$\text{We want to }subtract\text{ the sum of even numbers from 2 to 98: }S_2$

$\text{Note that: }\:S_2 \;=\;2 + 4 + 6 + \hdots + 98 \;=\;2(1 + 2 + 3 + \hdots + 49)$

$\text{Hence: }\:S_2 \:=\:2(\text{sum of first 49 numbers}) \;=\;2\left(\dfrac{49(49+1)}{2}\right)$

$\text{Therefore: }\:S \;=\;S_1 - S_2 \;=\;2\left(\dfrac{150(150+1)}{2}\right) - 2\left(\dfrac{49(49+1)}{2}\right)$

Thanks for that, so equally given the above could the subtration also read..

$\frac{300(300+1)}{2} - \frac{98(98+1)}{2}$
• Mar 5th 2011, 06:22 AM
emakarov
Quote:

Originally Posted by dumluck
Thanks for that, so equally given the above could the subtration also read..

$\frac{300(300+1)}{2} - \frac{98(98+1)}{2}$

No, $2\left(\dfrac{150(150+1)}{2}\right) - 2\left(\dfrac{49(49+1)}{2}\right)=\dfrac{300(150+1 )}{2}-\dfrac{98(49+1)}{2}$.
• Mar 5th 2011, 07:43 AM
dumluck
Thanks, I get why you have derived as above. My concern would be that I wouldn't recognise it in future questions. For example what if it was the same question but asked for the sum of odd numbers. I would assume the sum formula would be divisible by an odd number such as.. $\frac{n(n+2)}{3}$
• Mar 5th 2011, 08:27 AM
Wilmer
Sum odd numbers.
Code:

 1  3  5  7  9....  1  4  9 16 25....
See that?
• Mar 5th 2011, 08:42 AM
HallsofIvy
Quote:

Originally Posted by dumluck
Thanks, I get why you have derived as above. My concern would be that I wouldn't recognise it in future questions. For example what if it was the same question but asked for the sum of odd numbers. I would assume the sum formula would be divisible by an odd number such as.. $\frac{n(n+2)}{3}$

The sum of all integers from 1 to n is, as you noted in your first post, $\frac{n(n+1)}{2}$. Notice that the sum of even numbers, 2+ 4+ 6+ 8+ ...= 2(1+ 2+ 3+ 4+ ...) so that the sum of all even integers from 2 to 2n is 2 times the sum of all numbers from 1 to n: The sum of all even numbers from 1 to 2n is $n(n+ 1)$. The sum of odd numbers, from 1 to 2n+ 1, is the sum of all integers from 1 to 2n+1 minus the sum of all even numbers from 2 to 2n:
$\frac{2n(2n+1)}{2}- n(n+1)= \frac{2n(2n+1)}{2}- \frac{2n(n+1)}{2}= \frac{2n(2n+1- (n+1))}{2}$
and that gives the very simple formula Wilmer indicates.
• Mar 5th 2011, 09:18 AM
Soroban
Hello, dumluck!

It is surprising to learn that the sum of the first $\,n$ odd numbers is $\,n^2.$

There are various ways to prove this.
I enjoy this visual approach the most.

. . . . $\square$

. . . $1 \:=\:1^2$

. . . . $\begin{array}{c}
\square \blacksquare \\ [-1mm]
\blacksquare \blacksquare \end{array}$

. . $1 + 3 \;=\;2^2$

. . . . $\begin{array}{c}
\square \blacksquare \square \\ [-1mm]
\blacksquare \blacksquare \square \\ [-1mm]
\square \square \square \end{array}$

. $1 + 3 + 5 \;=\;3^2$

. . . . $\begin{array}{c}
\square \blacksquare \square \blacksquare \\ [-1mm]
\blacksquare \blacksquare \square \blacksquare \\ [-1mm]
\square \square \square \blacksquare \\ [-1mm]
\blacksquare \blacksquare \blacksquare \blacksquare \end{array}$

$1 + 3 + 5 + 7 \;=\;4^2$

• Mar 5th 2011, 10:08 AM
dumluck
Quote:

Originally Posted by Soroban
Hello, dumluck!

It is surprising to learn that the sum of the first $\,n$ odd numbers is $\,n^2.$

There are various ways to prove this.
I enjoy this visual approach the most.

. . . . $\square$

. . . $1 \:=\:1^2$

. . . . $\begin{array}{c}$ $
\square \blacksquare \\ [-1mm]
\blacksquare \blacksquare \end{array}" alt="
\square \blacksquare \\ [-1mm]
\blacksquare \blacksquare \end{array}" />

. . $1 + 3 \;=\;2^2$

. . . . $\begin{array}{c}$ $
\square \blacksquare \square \\ [-1mm]
\blacksquare \blacksquare \square \\ [-1mm]
\square \square \square \end{array}" alt="
\square \blacksquare \square \\ [-1mm]
\blacksquare \blacksquare \square \\ [-1mm]
\square \square \square \end{array}" />

. $1 + 3 + 5 \;=\;3^2$

. . . . $\begin{array}{c}$ $
\square \blacksquare \square \blacksquare \\ [-1mm]
\blacksquare \blacksquare \square \blacksquare \\ [-1mm]
\square \square \square \blacksquare \\ [-1mm]
\blacksquare \blacksquare \blacksquare \blacksquare \end{array}" alt="
\square \blacksquare \square \blacksquare \\ [-1mm]
\blacksquare \blacksquare \square \blacksquare \\ [-1mm]
\square \square \square \blacksquare \\ [-1mm]
\blacksquare \blacksquare \blacksquare \blacksquare \end{array}" />

$1 + 3 + 5 + 7 \;=\;4^2$

That's facinating so $1 + 3 + 5 + 7 + 9 + 11 = 6^2?$
• Mar 5th 2011, 10:23 AM
Wilmer
Quote:

Originally Posted by dumluck
That's facinating so $1 + 3 + 5 + 7 + 9 + 11 = 6^2?$

• Mar 5th 2011, 10:31 AM
dumluck
Quote:

Originally Posted by HallsofIvy
The sum of all integers from 1 to n is, as you noted in your first post, $\frac{n(n+1)}{2}$. Notice that the sum of even numbers, 2+ 4+ 6+ 8+ ...= 2(1+ 2+ 3+ 4+ ...) so that the sum of all even integers from 2 to 2n is 2 times the sum of all numbers from 1 to n: The sum of all even numbers from 1 to 2n is $n(n+ 1)$. The sum of odd numbers, from 1 to 2n+ 1, is the sum of all integers from 1 to 2n+1 minus the sum of all even numbers from 2 to 2n:
$\frac{2n(2n+1)}{2}- n(n+1)= \frac{2n(2n+1)}{2}- \frac{2n(n+1)}{2}= \frac{2n(2n+1- (n+1))}{2}$
and that gives the very simple formula Wilmer indicates.

thanks, let me try something here so I can further my understanding.

Let's say we are looking for the even numbers between 1 - 11
2+ 4+ 6+ 8+ 10 ...= 2(1+ 2+ 3+ 4+ 5 + 6 + 7 + 8 + 9 + 10) :

This gives us...
2+4+6+8+10 = 30
1+2+3+4+5+6+7+8+9+10 = 55 * 2 = 110.

so 30 != 110. So this is what is causing me some confusion.

1 3 5 7 9
1 4 9 16 25

The pattern here is as soraban states (i.e) $X^2$ but if we are looking for the odd integers in a range of 200 that is not starting at 0 would that not be quite difficult?

say odd integers from 800 - 1000.

Apologies for drawing this out, we're not all maths majors :).
• Mar 5th 2011, 10:34 AM
dumluck
It was a Rhetorical question. Just typing out loud.
Quote:

Originally Posted by Wilmer

. Sorry if I'm frustrating you.
• Mar 6th 2011, 02:44 AM
HallsofIvy
Quote:

Originally Posted by dumluck
thanks, let me try something here so I can further my understanding.

Let's say we are looking for the even numbers between 1 - 11

Do you mean the sum of the even numbers between 1 and 11?

Quote:

2+ 4+ 6+ 8+ 10 ...= 2(1+ 2+ 3+ 4+ 5 + 6 + 7 + 8 + 9 + 10) :
No, 2+ 4+ 6+ 8+ 10= 2(1+ 2+ 3+ 4+ 5) You were adding evens between 1 and 21, not 1 and 10.

Quote:

This gives us...
2+4+6+8+10 = 30
1+2+3+4+5+6+7+8+9+10 = 55 * 2 = 110.

so 30 != 110. So this is what is causing me some confusion.
You have misread what I wrote. I said before the sum of even numbers from 1 to 2n (NOT 1 to n) is n(n+ 1) which, for even numbers between 1 and 11, means 2n= 10 so n= 5. 5(6)= 30.

The sum of all even numbers between 1 and 21 has 2n= 20 so n= 10 and then 10(11)= 110.

Quote:

1 3 5 7 9
1 4 9 16 25

The pattern here is as soraban states (i.e) $X^2$ but if we are looking for the odd integers in a range of 200 that is not starting at 0 would that not be quite difficultsay odd integers from 800 - 1000. ?
Not a great deal. Find the sum from 1 to the ending number, the sum from 1 to the starting number, and subtract.

To find the the sum of all odd numbers between 800 and 1000 (801+ 803+ ...+ 997+ 999), we note that 999= 2(499)+ 1 so the sum of all odd numbers from 1 to 999 is $499^2= 249001$ and the sum of all odd numbers from 1 to 799= 2(399)+ 1 is $399^2= 159201$. The sum of all odd numbers between 800 and 1000 is 249001- 159201= 89800.

Quote:

Apologies for drawing this out, we're not all maths majors :).
I think Wilmer was "frustrated" at your refusal to add six integers yourself! You don't have to be a math major to do that.
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