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This is 7th grade pre-algebra math question. Please help......the teacher also wants to know what the sum of only the odd numbers would be i.e. 1+3+5+7......+495+497+499.

2006-09-12 09:57:28 · 13 answers · asked by Anonymous in Science & Mathematics Mathematics

13 answers

This is not as difficult as it sounds. You just need to break the problem into something easier to solve.

In this case, you can consider the numbers in pairs. 1+500=501, and 2+499 is also 501. If you pair up all the numbers, you can go all the way to the middle where 250 + 251 = 501. You know that this is the middle because the numbers are just one jump apart, and the first number tells you how many pairs there are. So the total is 250 pairs times the total for each pair:

250 x 501 = 125250

The same process works for your second problem. 1+499 = 500. 249+251 = 500. In this case we're going by twos, so you don't have 249 pairs, you have half as many... or 125 pairs (yah, half of 249 is 124.5, but don't forget you're starting from 1, not 0). Half as many pairs makes sense because you're counting by twos!

125 x 500 = 62500

Hope that helps!

2006-09-12 10:06:22 · answer #1 · answered by Doctor Why 7 · 2 1

''0+1 =' 1,''' 1/1 = 1,''''''''' 1*2=2, 2 - 1 = 1
''1+2 =' 3,''' 3/2 = 1.5, 1.5*2=3, 3 - 1 = 2
''3+3 =' 6,''' 6/3 = 2,''''''''' 2*2=4, 4 - 1 = 3
''6+4 =10, 10/4 = 2.5, 2.5*2=5, 5 - 1 = 4
10+5 =15, 15/5 = 3,''''''''''3*2=6, 6 - 1 = 5

hmm, works every time....so, using 500 and working backwards,

500+1=501, 501/2=250.5, 250.5*500=125,250
1+2+3+.....498+499+500 = 125,500

The second summation can be worked in much the same way. Don't forget to include the 0th term.

0+1=' 1 = 1*1, 1*2 = 2, 2-1=1
1+3=' 4 = 2*2, 2*2 = 4, 4-1=3
4+5=' 9 = 3*3, 3*2 = 6, 6-1=5
9+7=16 = 4*4, 4*2 = 8, 8-1=7

Eureka! Working backwards again,

499+1=500, 500/2=250, 250*250 = 62,500.
1+3+5+.....+495+497+499 = 62,500

Now, hide this, get out a sheet of paper & pencil, and see if you can duplicate it! Algebra makes figuring out things like this a whole lot easier!

2006-09-12 11:26:11 · answer #2 · answered by Helmut 7 · 0 0

This is a great problem. And it's attributed to a young Gauss.

Try arranging the numbers one time through least to greatest and the second time through from greatest to least.

1 + 2 + 3 + ... + 497 + 498 + 499
499+ 498+ 497+...3+ 2 + 1

(we've counted the numbers twice or double, i know.) Now add those two rows.

500 + 500 + 500+...+500+ 500+ 500 (how many 500s? well you had 499 numbers to count so 499 500s. 499*500 = 249500
and 249500 / 2 = 124750

Oops, I forgot to add 500. Well let's add it in! 124,750 + 500 = 125,250.

Now for 1 + 3 + 5 ... + 495 + 497 + 499.
499+ 497+ 495+...+5+3+ 1

The same 500+500+...+500 appears, but how many 500s this time is a harder question!
We had 499 before, which can't be split in half evenly. But notice that both the first number (1) and the last (499) are odd! So 250 odds exist!

250*500 / 2 = 62500

Now that you have the total of all number and the total from all the odds, consider what the total of the evens would be!

2006-09-12 10:08:41 · answer #3 · answered by J G 4 · 0 0

1 + 500 (first counting number plus last counting number) = 501

Multiply 501 by 250 because the 500 numbers can be divided into 250 groups of 2 (2 + 499, 3 + 498, . . . ).

The sum of the first 500 counting numbers is therefore 125,250, the product of 501 X 250

The sum of all odd numbers can be figured the same way; since 1 + 499 = 3 + 497 = 5 + 495 = . . ., you'll have groupings of 500

Counting only 250 odd numbers (half of 500), you'll have 125 groups that add up to 500. Figure out 500 X 125 = 62,500

So the sum of odd numbers 1 through 499 is 62,500

2006-09-12 10:43:44 · answer #4 · answered by ensign183 5 · 0 0

125250. If you pair up the first and last number of your series you get 1+500=501. Do that with the second and second to last terms and you get 2+499=501. If you keep pairing off the terms and adding them ( 3+498=501,4+497=501 etc.) you will have 250 such sums, the last one being 250+251=501. Since you have 250 sums each equaling 501 the sum of all the numbers in the series is 250x501=12520.

2006-09-12 10:42:31 · answer #5 · answered by True Blue 6 · 0 0

I know is a little too much for 7th grade but for the fun of algebra!
all(x)=(x^2+x)/2
all(500)=(500^2+500)/2=125250

even(x)=(x^2+2x)/4

odd(x)=(x^2+2x+1)/4
odd(499)=499^2+499*2+1)/4
=62500
it was so fun calculating the formulas!!!

2006-09-12 11:01:26 · answer #6 · answered by runlolarun 4 · 0 0

a = 1 ; { the first number }
l = 500 ; { the last number }
n =500 ; { how many numbers }
▪sum = n / 2 * ( a +l) , [ It's the formula you can use always]
sum = (500/2) * (1+500) = 250 * (501) = 125,250

Good Luck.

2006-09-12 10:14:43 · answer #7 · answered by sweetie 5 · 0 0

this is AP and we know that a=1, d=1 and n=500 in this case.

by formula, S500= 500/2 [2xa + ( n-1) d ]

S500= 250 { 2 + 499 }

S500= 250x 501

S500= 125, 250

2006-09-12 22:15:41 · answer #8 · answered by free aung san su kyi forthwith 2 · 0 0

Put it into an equation form: ( X+1 ) divided by X times 499= ???
Then ??? Devided by 2 = only odd numbers.

2006-09-12 10:10:41 · answer #9 · answered by joey 2 · 0 2

Sum of the first 500 counting numbers

125250

Sum of the odd numbers

62500

2006-09-12 10:02:54 · answer #10 · answered by David P 3 · 1 2

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