Question

The fundamental period of the function $f\left(x\right)=\left[x\right]+\left[2x\right]+\left[3x\right]+\cdots +\left[nx\right]-\frac{n(n+1)}{2}x,n\in N,$ is $($ where $[.]$ is the greatest integer function $)$

Answer 1

$f\left(x\right)=\left[x\right]+\left[2x\right]+\left[3x\right]+\cdots +\left[nx\right]-\frac{n(n+1)}{2}x,$
$\because \left[x\right]+\left\{x\right\}=x$
$f\left(x\right)=x-\left\{x\right\}+2x-\left\{2x\right\}+3x-\left\{3x\right\}+\cdots +nx-\left\{nx\right\}-\frac{n(n+1)}{2}x$
$f\left(x\right)=x(1+2+3+\cdots +n)-\frac{n(n+1)x}{2}-\left(\right\{x\}+\{2x\}+\{3x\}+\cdots +\{nx\left\}\right)$
$f\left(x\right)=x\frac{n(n+1)}{2}-\frac{n(n+1)}{2}x-\left(\right\{x\}+\{2x\}+\{3x\}+\cdots +\{nx\left\}\right)$
$f\left(x\right)=-\left(\right\{x\}+\{2x\}+\{3x\}+\cdots +\{nx\left\}\right)$

We know that period of $\left\{x\right\}$ is $1$
Hence, period of $f\left(x\right)$ is
$L.C.M.(1,\frac{1}{2},\frac{1}{3},\dots ,\frac{1}{n})$
$=\frac{L.C.M.(1,1,\dots ,n\text{times})}{H.C.F.(1,2,3,\dots ,n)}=1$