tex]...

Questions

August 2022 1 13 Report

Challenge for math lovers:
Find the general formula for the following sequence:
[tex] 5, 7, 10, 14, 15, 21, 20, 28, 25, 35, 30, 42, ... [/tex]

Answers & Comments

devoured

The general formula of the sequence is:

$\boxed{\displaystyle a_n=6\bigg\lceil\frac{n}{2}\bigg\rceil+(-1)^n\bigg\lceil\frac{n}{2}\bigg\rceil}$

(using ceiling function)

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Let $a_n$ to be the general term of the sequence. The rule we apply for the terms alternates between odd and even places; when the index $n$ is odd, $a_n$ will be multiple of 5, and when the index $n$ is even, $a_n$ will be a multiple of 7.

Case 1: When $n$ is odd.

If $n$ is odd, then we can set $n=2k-1$ for some positive integer $k.$ It follows that $n$ is the $kth$ positive odd integer. It can be observed that the terms in odd places in the sequence are the $kth$ multiple of 5. Therefore $a_n=5k$

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Case 2: When $n$ is even.

Same approach. If $n$ is even, then we can set $n=2k$ for some positive integer $k.$ It follows that $n$ is the $kth$ positive even integer. In this case, the terms in even places are the $kth$ multiple of 7 so $a_n=7k.$

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Combining the results, we can get that

$a_n=\begin{cases}5k \quad \textsf{if} \ n \ \textsf{odd}, \textsf{where} \ n=2k-1 \\ 7k \quad \textsf{if} \ n \ \textsf{even}, \textsf{where} \ n=2k\end{cases}$

$a_n=\begin{cases} 6k-k \ \ \textsf{if} \ n \ \textsf{odd}, \textsf{where} \ n=2k-1 \\ 6k+k \ \ \textsf{if} \ n \ \textsf{even}, \textsf{where} \ n=2k\end{cases}$

$a_n=6k\pm k$

The sign becomes negative if $n$ is odd and becomes positive if $n$ is even. Because the signs are alternating between even and odd, we can just write those as a power of -1;

$a_n=6k+(-1)^nk$

Note that the general term must be in terms of $n.$ We know that $n=2k-1$ if $n$ is odd and $n=2k$ if $n$ is even, solving for $k$ on each, we have

$\begin{array}{ccc} \displaystyle k=\frac{n+1}{2}=\frac{n}{2}+\frac{1}{2} & & \textsf{when} \ n \ \textsf{odd} \\ \\ \displaystyle k=\frac{n}{2} & & \textsf{when} \ n \ \textsf{even} \end{array}$

We need to make the values of $k$ on both cases to be equal in order to generalize the value of $k$ . To eliminate 1/2 on the case when $n$ is odd, we will use the ceiling function. One of the properties of the ceiling function is:

$\lceil x+n\rceil = \lceil x\rceil \quad \quad \textsf{if} \ 0\leq n$

Applying the ceiling function, we get

$\begin{array}{ccc} \displaystyle k=\bigg\lceil \frac{n}{2}+\frac{1}{2}\bigg\rceil=\bigg\lceil\frac{n}{2}\bigg\rceil & & \textsf{when} \ n \ \textsf{odd} \\ \\ \displaystyle k=\bigg\lceil \frac{n}{2}\bigg \rceil=\bigg \lceil \frac{n}{2}+0\bigg \rceil=\bigg\lceil\frac{n}{2}\bigg\rceil & & \textsf{when} \ n \ \textsf{even} \end{array}$