MHB Find Nth Term of Sequence: (-1)^n

[scottshannon]

I have been trying to find the nth term of the following sequence:
-1, 5, -17, 65...

I am thinking that the nth term has (-1) ^n . It doesn't appear to be a geometric or arithmetic sequence. I am stymied.

 

[alyafey22]

I have been trying to find the nth term of the following sequence:
-1, 5, -17, 65...

I am thinking that the nth term has (-1) ^n . It doesn't appear to be a geometric or arithmetic sequence. I am stymied.

It seems if $a_n = (-1)^{n-1} (4^{n}+1) \,\,\,\, n\geq 1$

 

[skeeter]

It seems if $a_n = (-1)^{\color{red}{n}} (4^{\color{red}{n-1}}+1) \,\,\,\, n\geq 1$

correction

 

[scottshannon]

Thank you very much

 

[alyafey22]

correction

why ? If $n=1$ then you get $a_1 = -2$.

 

[scottshannon]

yes you are correct..i overlooked that...

 

[scottshannon]

I wonder if anyone might have any further ideas? This nth term formulation gives all terms except the first.

 

[Evgeny.Makarov]

I have been trying to find the nth term of the following sequence:
-1, 5, -17, 65...

If the first number has index 1, then these numbers are described by the following formula.
\[
a_n=
\begin{cases}-1&n=1\\
(-1)^n (4^{n-1}+1)&n>1
\end{cases}\qquad(*)
\]
The sequence (*) is also described by infinitely many other formulas. Also, there are infinitely many sequences that start with -1, 5, -17, 65…, and each is described by infinitely many formulas.

 

[scottshannon]

"Also, there are infinitely many sequences that start with -1, 5, -17, 65…, and each is described by infinitely many formulas. "

Wow..I had no idea. There are a couple of things that had never occurred to me:

1) That it is permissible to write the nth term this way.

2) Also that there might be infinitely many nth term representations for a series where only the first 4 terms are given. Is it valid then to say that as one sees more terms in the series that the number of possibilities for the nth term decreases? If it is true that there are an infinite number of ways of writing the nth term representation, then will every one of them give the same 5th, 6th...terms?

 

[Evgeny.Makarov]

There are a couple of things that had never occurred to me:

1) That it is permissible to write the nth term this way.

It is permissible in principle. So without any restrictions this problem is not interesting. One could write a formula
\[
a_n=
\begin{cases}
-1&n=1\\
5&n=2\\
-17&n=3\\
65&n=4\\
a_n=1&n>4.
\end{cases}
\]
It specifies a valid function from natural numbers to natural numbers, and its values for $n=1,\dots,4$ are as required.

A specific problem may impose restrictions on how the function (i.e., the sequence) is specified. For example, it is possible to require that the expression specifying $a_n$ uses only addition, subtraction, multiplication and division. Then one solution is $a_n=22 n^3-146 n^2+290 n-167$ (it does not even use division). You can verify that $a_n$ are as required for $n=1,\dots,4$.

It is also possible to ask for a shortest expression. This problem may be interesting, but I haven't seen something like it.

Also that there might be infinitely many nth term representations for a series where only the first 4 terms are given.

One can do something silly such as changing an expression $E$ to $E+1+\dots+1-1-\dots-1$ where all 1's cancel. Strictly speaking, this expression specifies the same function, but it is a different from $E$. But one can also use write expressions such that proving their equality to $(-1)^{n} (4^{n-1}+1)$ would require highly nontrivial mathematics.

Is it valid then to say that as one sees more terms in the series that the number of possibilities for the nth term decreases?

In some sense, but it is still infinite. An infinite set $A$ may be a subset of another set $B$ and yet have as many elements as $B$.

If it is true that there are an infinite number of ways of writing the nth term representation, then will every one of them give the same 5th, 6th...terms?

For each sequence (function) there is an infinite number of formulas (expressions) that specify it. Also, there are infinitely many sequences with a given initial finite segment such as -1, 5, -17, 65.

 

[scottshannon]

It is permissible in principle. So without any restrictions this problem is not interesting. One could write a formula
\[
a_n=
\begin{cases}
-1&n=1\\
5&n=2\\
-17&n=3\\
65&n=4\\
a_n=1&n>4.
\end{cases}
\]
It specifies a valid function from natural numbers to natural numbers, and its values for $n=1,\dots,4$ are as required.

A specific problem may impose restrictions on how the function (i.e., the sequence) is specified. For example, it is possible to require that the expression specifying $a_n$ uses only addition, subtraction, multiplication and division. Then one solution is $a_n=22 n^3-146 n^2+290 n-167$ (it does not even use division). You can verify that $a_n$ are as required for $n=1,\dots,4$.

It is also possible to ask for a shortest expression. This problem may be interesting, but I haven't seen something like it.

One can do something silly such as changing an expression $E$ to $E+1+\dots+1-1-\dots-1$ where all 1's cancel. Strictly speaking, this expression specifies the same function, but it is a different from $E$. But one can also use write expressions such that proving their equality to $(-1)^{n} (4^{n-1}+1)$ would require highly nontrivial mathematics.

In some sense, but it is still infinite. An infinite set $A$ may be a subset of another set $B$ and yet have as many elements as $B$.

For each sequence (function) there is an infinite number of formulas (expressions) that specify it. Also, there are infinitely many sequences with a given initial finite segment such as -1, 5, -17, 65.

Thank you very much