â„–4284826[Quote]
(n^3 - n)/6
â„–4284832[Quote]
How did you manage to butcher basic math lmao
â„–4284837[Quote]
sum(i=0,n) i(i-1)/2=f(n)
f(n)=1/2 (sum(i=0,n) i^2)-1/2 sum(i=0,n) i
f(n)=1/2 (sum(i=0,n) i^2)-(n^2-n)/4
you can find out what sum i ^2 is by using the same method used to find what sum i is on sum i^2
â„–4284869[Quote]
>If 1+2+3+4+5+6+7+…+n Is equal to doing (n(n-1))/2
How is it equal?
Hang on I am testing this
â„–4284879[Quote]
>>4284801 (OP)Nigger it isnt equal
â„–4284881[Quote]
>>4284869He is wrong. 1+2+3+..+n is actually equal to (n(n+1))/2
â„–4284882[Quote]
>>4284879Its actually n(n+1)/2 that is equal
â„–4284885[Quote]
.
â„–4284970[Quote]
>>4284837i am so retarded
â„–4284977[Quote]
>>4284947I am kind of curious on what this may be now
â„–4284998[Quote]
>>4284947Janniefaggot is probably trying to understand my solution, fucking wanker
â„–4285040[Quote]
>>4284947Up. Every /pol/sir must see this