Havishma Haranath
Last Activity: 9 Years ago
I think the answer is onr.
We generally know that [x+integer]=[x]+integer
let’s consider f(x+1)= [x+1]+[2(x+1)]+....+[n(x+1)]-n(x+1)(n+1)/2
solving which we get f(x+1)=[x]+1+[2x]+2+....+[nx]+n-n(n+1)/2-nx(n+1)/2
since 1+2+3+...+n=n(n+1)/2
we get f(x+1)=f(x)
similarly we can also show that f(x)=f(x+1)=f(x+2)=..f(x+any integer)
but since period must be the least posibble number
period is !