# Thread: sum of first 1000 odd numbers is 100,000,000 but what is the

1. ## sum of first 1000 odd numbers is 100,000,000 but what is the

I'm trying to determine the formula for the sum of the first 1000 odd digits.
I know that the sum of all number from 1 to 1000 is n(n+1)/2 but I can't figure out the odd number formula. Please help!

2. Another way.
If you know that then:
The first number in an arithmetic progression is a and the difference is d :
y=a+(a+d)+(a+2d)+(a+3d)...+(a+(n-1)d) =
n*a+sum(from 1 to (n-1))*d =
n*a+ (d*(n-1)*n/2) =
(2a + d*(n-1))*n/2

3. Surprisingly, the sum of the first n odd integers is n^2.

How do you find the sum of a series of odd or even numbers?

Consider the odd numbers.

n......1......2......3......4......5......6
N.....1......3......5......7......9.....11
Sum.1.....4......9.....16.....25....35 (Look familiar?)
Diff......3......5......7......9......11
Diff.........2.......2......2......2

With the 2nd differences constant, the sequence of sums forms a fintite difference sequence.
The general expression for the nth term (the sum of n odd numbers) is of the form an^2 + bn + c.
Using the given data, we can write

a(1^2) + b(1) + c = 1
a(2^2) + b(2) + c = 4
a(3^2) + b(3) + c = 9

Solving, a = 1, b = 0 and c = 0.
Therefore, the general expression for the sum of "n" consecutive odd numbers starting with 1 is S(n) = n^2.

Consider the even numbers.

n......1......2......3......4......5......6
N.....2......4......6......8.....10....12
Sum2......6.....12....20.....30....42
Diff.....4.......6......8.....10.....12
Diff..........2......2......2......2

With the 2nd differences constant, the sequence of sums forms a fintite difference sequence.
The general expression for the nth term (the sum of n odd numbers) is of the form an^2 + bn + c.
Using the given data, we can write

a(1^2) + b(1) + c = 2
a(2^2) + b(2) + c = 6
a(3^2) + b(3) + c = 12

Solving, a = 1, b = 1 and c = 0.

Therefore, the general expression for the sum of "n" consecutive even numbers starting with 2 is S(n) = n^2 + n = n(n + 1).

4. The sum of the first n odd numbers is n^2

n....1....2.....3.....4
......1 = 1
......1 + 3 = 4
......1 + 3 + 5 = 9
......1 + 3 + 5 + 7 + 9 = 25, etc.

* The nth square is equal to n^2.
n....1....2....3....4....5....6....7....8....9.... 10
Sq..1...4....9....16..25..36..49..64...81...100

The sum of the first 1000 odd numbers is 1000^2 = 1,000,000.

5. ## Re: sum of first 1000 odd numbers is 100,000,000 but what is

Hello, aciokajlo!

. . It's primitive and quite long, but it's nice to have options, isn't it?

]I'm trying to determine the formula for the sum of the first 1000 odd digits.
I know that the sum of all number from 1 to 1000 is n(n+1)/2
but I can't figure out the odd number formula.
You already know the sum of all the numbers from 1 to 1000, right?
. . . (1000)(1001)/2 .= .500,500

Now find the sum of the even numbers from 2 to 1000.
. . . 2 + 4 + 6 + ... + 1000 .= .2(1 + 2 + 3 + ... + 500)

This is: . 2 x (sum of 1 to 500) .= .2 x (500)(501)/2 .= .250,500

Therefore: .(sum of odd) .= .(sum of all) - (sum of even) .= .500,500 - 250,500 .= .250,000

. . See? . . . You <u>could</u> have figured it out yourself!

6. To repeat:

Surprisingly, the sum of the first n odd integers is simply n^2.

The first odd number is 1. The 1000th odd number is 2(1000) - 1 = 1999.

How do you find the sum of a series of odd numbers?

Consider the odd numbers.

n..........1......2......3......4......5......6... ...7......8......9......10.............n
Odd#...1......3......5......7......9.....11.....13 .....15....17.....19 .........2n-1
Sum.....1.....4......9.....16.....25....36........ ....................................n^2

Notice anything? The sum of any consecutive sequence of odd numbers starting with 1 is n^2.

I gave you the analytical method earlier.

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