# Thread: Proof of [x]+[(x+1)/n]+....+[(x+n-1)/n]=[x],Kindly help me in this

1. ## Proof of [x]+[(x+1)/n]+....+[(x+n-1)/n]=[x],Kindly help me in this

Proof of [x]+[(x+1)/n]+....+[(x+n-1)/n]=[x],Kindly help me in this.

2. ## Re: Proof of [x]+[(x+1)/n]+....+[(x+n-1)/n]=[x],Kindly help me in this Originally Posted by vivek 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  ?)

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

3. ## Re: Proof of [x]+[(x+1)/n]+....+[(x+n-1)/n]=[x],Kindly help me in this

Its my bad, on L.H.S , it was [x/n] instead of [x],and [x] is greatest integer function

4. ## Re: Proof of [x]+[(x+1)/n]+....+[(x+n-1)/n]=[x],Kindly help me in this Originally Posted by topsquark 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  ?)

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.
Lets consider $P(n,x)]=\sum\limits_{k = 0}^{n - 1} {\left\lfloor {\dfrac{{x + k}}{n}} \right\rfloor }$

Now look at $P(10,4)$

5. ## Re: Proof of [x]+[(x+1)/n]+....+[(x+n-1)/n]=[x],Kindly help me in this

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

6. ## Re: Proof of [x]+[(x+1)/n]+....+[(x+n-1)/n]=[x],Kindly help me in this

$x=nq+r$ 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

$$\left\lfloor\dfrac{x+k}{n}\right\rfloor=q+1$$