How to get this shit solved

consider the modulus function

>>8188040
Man I learned how to add to fractions together a month ago ,all I know is the sum formula for the arithmetic and geometric sequences ,tried some logic and on the arithmetic sum formula and I've got n^2 for the even sum, where n is the number of values if I plug in 5 it will sum up 5 values from the series.

>>8188032
Is this how your sequence works?
n=0: {0}
n=1: {0,1,2}
n=2: {0,1,2,3,4}
n=3: {0,1,2,3,4,5,6}
n=4: {0,1,2,3,4,5,6,7,8}
...
If so then it took me like 5minutes to figure this out: the sum of the odd numbers is n^2 and the sum of the even numbers is n^2+n.
Just take the derivative of x^2 or play around with triangle numbers or look for patterns in a times table, you'll understand how I got those equations.

>>8188054
The even and odd sum is supposed to be the index where the numbers are for example 0,1,2,3,4,5,...2n
1 is at index 2
3 is at index 4
And so on

>>8188054
Look at it like a normal arithmetic sequence

>>8188032
Take the average and multiply it by the number of terms. Average of the evens is n. There are n+1 evens. So n(n+1) is the sum of evens. Average of the odds is also n. There are n odds. So n^2 is the sum of odds. Brainlet

File: 14678427328181542881308.jpg (2MB, 4160x3120px) Image search: [Google]

2MB, 4160x3120px

>>8188070
Here is what I've tried ,I'm not trying to sum up the even numbers I'm trying to sum up the numbers at the even places first one is 0 its at the odd place ,second number is 1 so its at an even place.
What you are seeing in the pic is the arithmetic sum formula.

>>8188088
Don't you realize that its exactly the same except that evens are at odd places and odds are at even places? Why not just say sum the odds numbers and even numbers?

>>8188122
Cause I'm an idiot apperantly

>>8188032
>>8187214

stop making threads you fucking pleb

These are called triangle numbers except your index is increased by 1. Google it for a easy way to visualize.

We know 1 + 2 + 3 + ... + n = n(n+1)/2.
Therefore 2 + 4 + 6 + ... + 2n = n(n+1),
and 1 + 3 + 5 + ... + (2n -1) = 2n(2n+1)/2 - n(n+1).