A problem of seconds in seconds

Algebra Level 3

$\large \sum_{z=1}^{2016} r_{z}=?$ where $r_{z}$ is the $z^{th}$ term of a sequence of distinct real numbers such that $r_1=0$ , $r_{2016}=2$ and $|r_{n+1}-r_n|$ is constant.

The answer is 2016.

1 solution

Zeeshan Ali
Jan 14, 2016

We have $n=2016$ , $r_{1}=0$ , $r_{2016}=2$ and $|r_{n+1}-r_n|$ is constant. Simply, they form an Arithmetic Progression. Since $S_{n}=\frac{n}{2} \times \left( r_{1}+r_{n} \right)$ , therefore: $S_{2016}=\frac{2016}{2} \times \left( 0+2 \right)=\frac{2016}{2} \times \left( 2 \right)=\boxed{2016} :)$

Moreover, if you wanna find out the difference between $T_n$ and $T_{n+1}$ , just follow the relation that $T_n=T_1+(n-1)\times d$ where n, $T_1$ , $T_n$ and d be the total number of values, $1^{st}$ term, $n^{th}$ term and difference of an A.P. respectively.

Here n=2016, $T_1=0$ and $T_n=2$ . Therefore: $2=0+(2016-1) \times d \implies d=\frac{2}{2015}$

Hence the numbers are equidistant by a difference of $\boxed{d=\frac{2}{2015}}$ .

Please explain the first line in more detail. Thanks!

Calvin Lin Staff - 5 years, 5 months ago

As we can find out the constant different $d=|r_{n+1}-r_{n}|= \frac{2}{2015}$ .

Hence the series has unique values as under: $0, 0+ \frac{2}{2015}, \frac{2}{2015}+ \frac{2}{2015}, \frac{4}{2015}+ \frac{2}{2015},...,2$

Zeeshan Ali - 5 years, 5 months ago

Not quite. As previously stated, you have to carefully untangle the convoluted phrasing, to understand what the setup is, and then proving that we have found all possible sequences. It is not sufficient to merely state a sequence that satisfies the conditions which you have laid out. If you do so, then you are then making an (unproven) assumption that there is a unique sequence which satisfies the conditions.

Here is what I would do:

First, because we have distinct values, even though $|r_{n+1} - r_n| = |r_{n-1} - r_n |$ , we do not have $r_{n-1} = r_{n+1}$ .
Second, since $|x - r_n| = c$ has 2 unique solutions, namely $r_n \pm c$ , thus if $r_{n-1}$ and $r_n$ are determined, then so is $r_{n+1}$ .
Third, use this to show that $r_i$ is arranged in order.
Fourth, since $r_1 = 1, r_{2016} = 2$ , hence this sequence is in increasing order.
Fifth, since the difference is a constant, and we have an increasing sequence, thus we must have an arithmetic progression.

Calvin Lin Staff - 5 years, 5 months ago

A problem of seconds in seconds

The answer is 2016.

1 solution

0 pending reports