# Thread: Find the sum, arithmetic sequence

1. ## Find the sum, arithmetic sequence

determine the difference between the sum of the 1000 first even positive integers and the sum of the first 1000 odd positive integers.

this is how I thought:

2+4+6+8+...+1000

an=a1+(n-1)*d

1000=2+(n-1)*2
1000=2+2n-2
1000-2+2=2n
n=500

sum=(n/2)*(a1+an)=(500/2)*(2+1000)
sum=250500

Then I tought

1+3+5+7+...+1000

an=a1+(n-1)*d
1000=1+(n-1)*2
1000=1+2n-2
1000=-1+2n
1001=2n
n=500,5

sum=(500,5/2)*(1+1000)
sum=250500,25

250500,25-
250500=0,25
The answer is 1000. How did I Count wrong??

2. ## Re: Find the sum, arithmetic sequence Originally Posted by selin determine the difference between the sum of the 1000 first even positive integers and the sum of the first 1000 odd positive integers.
this is how I thought:

2+4+6+8+...+1000
an=a1+(n-1)*d
1000=2+(n-1)*2
1000=2+2n-2
1000-2+2=2n
n=500

sum=(n/2)*(a1+an)=(500/2)*(2+1000)
sum=250500

Then I tought

1+3+5+7+...+1000

an=a1+(n-1)*d
1000=1+(n-1)*2
1000=1+2n-2
1000=-1+2n
1001=2n
n=500,5

sum=(500,5/2)*(1+1000)
sum=250500,25

250500,25-
250500=0,25
The answer is 1000. How did I Count wrong??
$\sum\limits_{k = 1}^{{{10}^3}} {(2k)} = 2\sum\limits_{k = 1}^{{{10}^3}} {(k)} = ?$

$\sum\limits_{k = 1}^{{{10}^3}} {(2k - 1)} = 2\sum\limits_{k = 1}^{{{10}^3}} {(k)} - \sum\limits_{k = 1}^{{{10}^3}} {(1)} = ?$

3. ## Re: Find the sum, arithmetic sequence Originally Posted by selin determine the difference between the sum of the 1000 first even positive integers and the sum of the first 1000 odd positive integers.

this is how I thought:

2+4+6+8+...+1000 No. This is only the first 500 even numbers.

an=a1+(n-1)*d

1000=2+(n-1)*2
1000=2+2n-2
1000-2+2=2n
n=500 No, you are told n=1000

sum=(n/2)*(a1+an)=(500/2)*(2+1000)
Use sum = (n/2)(a1 + (n-1)d) where n=1000, a=2 and d=2
sum=250500

Then I tought

1+3+5+7+...+1000
No. 1000 is not odd.

an=a1+(n-1)*d
1000=1+(n-1)*2
1000=1+2n-2
1000=-1+2n
1001=2n
n=500,5 Firstly n must be an integer. Secondly you are told that n=1000

sum=(500,5/2)*(1+1000)
sum=250500,25 How could this not be an integer?? Use sum = (n/2)(a1+(n-1)d) where a=1, n=1000, d=2

250500,25-
250500=0,25
The answer is 1000. How did I Count wrong??

If you make the changes I suggested you will get the correct answer. But that's not the most efficient way of finding the solution to the problem.

Think of it this way:
Sum of first 1000 odd numbers = 1 + 3 + 5 + 7 + . until you have 1000 terms
Sum of first 1000 even numbers = 2 + 4 + 6 + 8 +  until you have 1000 terms = (1+1)+(3+1)+(5+1)+(7+1) +  until you have 1000 pairs

Notice that each term in the "even" sum is 1 more than the corresponding term in the "odd" sum.

So the "even" sum will be 1000 x 1 = 1000 more than the "odd" sum.

4. ## Re: Find the sum, arithmetic sequence Originally Posted by selin determine the difference between the sum of the 1000 first even positive integers and the sum of the first 1000 odd positive integers.
The answer is 1000. How did I Count wrong??
Do you know the so-called Gauss-Sum : $\sum\limits_{k = 1}^N k = \frac{{N(N + 1)}}{2}$ that is miss-named but that is another story.
So $\sum\limits_{k = 1}^N {(2k)} = 2\sum\limits_{k = 1}^N {(k)} = 2\frac{{N(N + 1)}}{2} = N(N + 1)$

If $\sum\limits_{k = 1}^N {(2k)}$ is the sum of the first $\displaystyle N$ even integers.
Each number in that sum of even integers is $2k$.

So look at the sum of the first $\displaystyle N$ odd integers $\displaystyle \sum\limits_{k = 1}^N {(2k -1) } = \sum\limits_{k = 1}^N {(2k) }-1 \sum\limits_{k = 1}^N {(1)}$

Can you see that each odd number in that range is on less that the corresponding even number? There are $\displaystyle 10^3$ such numbers.
SO?

5. ## Re: Find the sum, arithmetic sequence

My post #3 is the layman's version of what Plato just said.

6. ## Re: Find the sum, arithmetic sequence

I haven't learned what that E means. But I still managed to solve it, thank you anyway.

7. ## Re: Find the sum, arithmetic sequence

solved it! thanks!