SOLUTION: Let n=1+2+3+5+.....+999 and m=2+4+6+....+1000. Then m-n equals ?

Question 629901: Let n=1+2+3+5+.....+999 and m=2+4+6+....+1000. Then m-n equals ?
Answer by htmentor(1343) (Show Source): You can put this solution on YOUR website!
[I assume you mean n=1+3+5+..., in other words only odd numbers]
We can do this problem formally by determining the sums of arithmetic sequences, etc.
But if we recognize that each time another term is added to the sequences, the sum of m grows larger by 1 compared to n.
For example, the sum of the first 3 terms is 9 and 12.
Since we've added 3 terms sum m is larger by 3*1 = 3
Since the sequences have 500 terms, subtracting the two sums gives 500*1 = 500
If you'd like the formal solution, here it is:
The n-the term of an arithmetic sequence can be written a(n) = a + (n-1)d, where a=the 1st term, d=the common difference
In both cases, the common difference is 2, so we can write
a(n) = 2n - 1 [for sequence n]
a(n) = 2n [for sequence m]
The sum of the 1st n terms of an arithmetic sequence = (n/2)(a+a(n))
The sum of n = 250(1+999) = 250000
The sum of m = 250(2+1000) = 250500
We see the difference is 500, as we determined above.

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