SOLUTION: without computing each sum, find which is greater,O or E. O=5+7+9+...+99 and E=4+6+8+10+...+100?

Question 1163462: without computing each sum, find which is greater,O or E. O=5+7+9+...+99 and E=4+6+8+10+...+100?
Found 2 solutions by greenestamps, Boreal:
Answer by greenestamps(13203) (Show Source): You can put this solution on YOUR website!

Remove 4 from E and you have

O = 5+7+9+...+97+99
E = 6+8+10+...+98+100

In those two sums, the numbers of terms are the same, and each term of E is 1 more than the corresponding term of O.

So even without the term 4 in the given definition of E, E is greater than O. So certainly with the additional term 4 in E, E is greater than O.

Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
There are 100 potential integers, 50 even and 50 odd.
Two of the odd are not counted (1 and 3)
One of the even is not counted (2).
The way to line them up is:
4+6+8+10+...+100. there are 49 even numbers in the series
---5+7+9+...+99. There are 48 odd numbers in the series
The even is greater with a larger number for the last 48 in the series and a leading number of 4 as well.
There are 48 pairs where E is 1 greater and a 4 that is not paired, so 52 greater.
if you want the sum (not the question)
(n/2)*(2a+(n-1)d)
(49/2)*(8+48*2)=2548 for the even
(48/2)*(10+47*2)=2496 for the odd
They sum to 5044
add the missing numbers 1,2,3 and the sum is 5050, the sum of the first hundred integers
n(n+1)/2=100*101/2=5050

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