Proof of [x]+[(x+1)/n]+....+[(x+n-1)/n]=[x],Kindly help me in this.
What does [x] mean?
Either way, we can cancel the [x]'s on each side, leading to
$\displaystyle \left [ \frac{x + 1}{n} \right ] + \text{ ... } + \left [ \frac{x + n - 1}{n} \right ] = 0$
(Or perhaps [0] ?)
Let's say that n = 4. Then we have
$\displaystyle \left [ \frac{x + 1}{4} \right ] + \left [ \frac{x + 2}{4} \right ] + \left [ \frac{x + 3}{4} \right ] = 0$
This doesn't seem right.
-Dan
first assume $\displaystyle 0\leq x<n$
then show that
$\displaystyle \left\lfloor \frac{x+k}{n}\right\rfloor =$
$\displaystyle 0$ if $\displaystyle 0\leq k\leq n-\lfloor x\rfloor -1$
$\displaystyle 1$ otherwise
$x=nq+r$ where $0\le r < n$
So, $$\left\lfloor \dfrac{x}{n}\right\rfloor =q$$ but $$\left\lfloor \dfrac{x+n}{n}\right\rfloor = q+1$$
For all $0\le k <n-r$, you have:
$$\left\lfloor\dfrac{x+k}{n}\right\rfloor=q$$
For all $n-r \le k < n$, you have
$$\left\lfloor\dfrac{x+k}{n}\right\rfloor=q+1$$