Sum of some floored and summed series

Algebra Level 5

Let a series be defined such that

$t_{r+1}= \displaystyle \sum_{m=0}^r \left\lfloor \frac{(x+m)}{r+1} \right\rfloor .$

Let another series be defined as $T_{r}= \left \lfloor t_{r}+r^{2} \right \rfloor .$

if $\displaystyle \sum_{r=1}^n T_{r}$ can be written as

$n \left \lfloor \frac{1}{n}+(n+1)(2n+1) \right \rfloor - \frac{5}{6}n(n+1)(2n+1) ,$

find the value of $\Large \color{#D61F06}{x}$ .

Given that $n=100$

Details and Assumptions

among the infinite values of $x$ , the value asked for is also defined as one hundredth of the number of integral values of $x$

2 solutions

Rohith M.Athreya
Mar 11, 2016

$t_{r+1}= \sum_{m=0}^r \lfloor \frac{(x+m)}{r+1} \rfloor$ is a constant value ie., $\lfloor x \rfloor$

$r^{2}$ is a natural number

thus, $\lfloor t_{r}+r^{2} \rfloor$ is $\lfloor t_{r} \rfloor + r^{2}$

thus,summation results in,

$n\lfloor x \rfloor+\frac{n(n+1)(2n+1)}{6}$

$n\lfloor x+(n+1)(2n+1) \rfloor - \frac{5n(n+1)(2n+1)}{6}$ ------------ $\boxed{i}$

$\lfloor x \rfloor =0$ now the number of integral values satisfying this is 1 ie.,0. thus, $x$ is 0.01

Refer : Properties of Floor Functions

But if we substitute $x=0.02$ ,that also satisfies the problem.

Arihant Samar - 5 years, 3 months ago

it works ,but the question states that $\boxed{i}$ can be expressed in a certain manner ie.,

$n \left \lfloor \frac{1}{n}+(n+1)(2n+1) \right \rfloor - \frac{5}{6}n(n+1)(2n+1) ,$

if $x$ is 0.02, $\frac{1}{n}$ will not be $x$

despite there being infinite solutions to the equation, the specified form in question limits the answers to $\boxed{0.01}$

Rohith M.Athreya - 5 years, 3 months ago

All values from (-1; 1) are solutions how a numerical analysis via XPLORE shows meaning there isn't any unique solution for x demanded in this task!

Andreas Wendler - 5 years, 3 months ago

@Andreas Wendler – made an addition to the question in order to get a unique solution

Rohith M.Athreya - 5 years, 3 months ago

@Rohith M.Athreya – Yes, you are right. But nevertheless the task seems to be a bit quirky ;-( !

Andreas Wendler - 5 years, 3 months ago

Danish Juneja
Aug 6, 2020

I just used the property that [x] + [x + 1 / n] + [x + 2 / n].....[x + (n - 1 / n)] = [nx], Next was just simple rearrangement