# Convert the recursive formula into the explicit form

• MHB
• mathmari
In summary: Thinking)In summary, the sequence given is $0, 2, -6, 12, -20, \dots$ and its recursive definition is $a_1=0$ and $a_{n+1}=(-1)^{n+1}\cdot (a_n+2\cdot n)$. To convert it to an explicit form, we can first consider the sequence without the signs, which is $0, 2, 6, 12, 20, \dots$. We can then use telescoping to get the explicit formula $a_n=n(n-1)$. To account for the signs, we can add the $(-1)^{n+1}$ back to get the final
mathmari
Gold Member
MHB
Hey!

We have the sequence $$0, \ 2 , \ -6, \ 12, \ -20, \ \ldots$$ Its recursive definition is \begin{align*}&a_1=0 \\ &a_{n+1}=(-1)^{n+1}\cdot (a_n+2\cdot n)\end{align*} or not?
How can we convert that in the explicit form? (Wondering)

It is easiest first of all to ignore the $(-1)^{n+1}$ and consider the sequence
$$0,2,6,12,20,\ldots.$$
We can add the necessary minus signs later on.

This sequence is thus $a_{n+1}=a_n+2n$ i.e. $a_{n+1}-a_n=2n$. We can therefore use telescoping to get an explicit formula:
$$\begin{array}{rcl}a_{n+1}-a_n &=& 2n \\ a_n-a_{n-1} &=& 2(n-1) \\ {} &\vdots& {} \\ a_2-a_1 &=& 2\cdot1\end{array}$$
$\displaystyle\implies\ a_{n+1}-a_1=2\sum_{r=1}^nr=2\cdot\frac{n(n+1)}2=n(n+1)$,

i.e. $a_n=n(n-1)$. To restore the minus signs, simply add the $(-1)^{n+1}$ back:
$$\boxed{a_n\ =\ (-1)^{n+1}n(n-1)}.$$

PS: The formula for the original sequence $0,2,-6,12,-20,\ldots$ should be
$$a_1=0;\ a_{n+1}=a_n+(-1)^{n+1}\cdot2n.$$
If it were
mathmari said:
\begin{align*}&a_1=0 \\ &a_{n+1}=(-1)^{n+1}\cdot (a_n+2\cdot n)\end{align*}
the fourth term would be $0$, not 12.

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mathmari said:
Hey!

We have the sequence $$0, \ 2 , \ -6, \ 12, \ -20, \ \ldots$$ Its recursive definition is \begin{align*}&a_1=0 \\ &a_{n+1}=(-1)^{n+1}\cdot (a_n+2\cdot n)\end{align*} or not?
How can we convert that in the explicit form? (Wondering)

Olinguito said:
$$\boxed{a_n\ =\ (-1)^{n+1}n(n-1)}.$$

Hey mathmari and Olinguito!

Just an observation:
$$\begin{array}{c|c|c|c|}n & a_n & a_{n+1}=(-1)^{n+1}\cdot (a_n+2\cdot n) & a_n\ =\ (-1)^{n+1}n(n-1) \\ \hline 1 & 0 & 0 & 0\\ 2 & 2 & (-1)^{1+1}\cdot(0+2\cdot 1) = 2 & (-1)^{2+1}\cdot 2 (2-1)={\color{red}-2}\\ 3 & -6 & (-1)^{1+2}\cdot(2+2\cdot 2) = -6 \\ 4 & 12 & (-1)^{1+3}\cdot(-6+2\cdot 3) = {\color{red}0} \\ 5 & -20 \\ \end{array}$$
(Thinking)

I like Serena said:
ust an observation:
$$\begin{array}{c|c|c|c|}n & a_n & a_{n+1}=(-1)^{n+1}\cdot (a_n+2\cdot n) & a_n\ =\ (-1)^{n+1}n(n-1) \\ \hline 1 & 0 & 0 & 0\\ 2 & 2 & (-1)^{1+1}\cdot(0+2\cdot 1) = 2 & (-1)^{2+1}\cdot 2 (2-1)={\color{red}-2}\\ 3 & -6 & (-1)^{1+2}\cdot(2+2\cdot 2) = -6 \\ 4 & 12 & (-1)^{1+3}\cdot(-6+2\cdot 3) = {\color{red}0} \\ 5 & -20 \\ \end{array}$$
(Thinking)

Ah should it maybe be $$a_{n+1}=(-1)^{n+1}\cdot (|a_n|+2\cdot n)$$ ?

Because if we consider the sequence without the signs, we add at the previous number the number 2n. Then the sign changes, at the odd positions we have $-$ and at the even places we have $+$, or not?

(Wondering)

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I like Serena said:
$$\begin{array}{c|c|c|c|}n & a_n & a_{n+1}=(-1)^{n+1}\cdot (a_n+2\cdot n) & a_n\ =\ (-1)^{n+1}n(n-1) \\ \hline 1 & 0 & 0 & 0\\ 2 & 2 & (-1)^{1+1}\cdot(0+2\cdot 1) = 2 & (-1)^{2+1}\cdot 2 (2-1)={\color{red}-2}\\ 3 & -6 & (-1)^{1+2}\cdot(2+2\cdot 2) = -6 \\ 4 & 12 & (-1)^{1+3}\cdot(-6+2\cdot 3) = {\color{red}0} \\ 5 & -20 \\ \end{array}$$
Thanks, ILS. When the first term is $0$ it’s easy to slip into thinking it’s $a_0$ rather than $a_1$. In this case the formula should be
$$\boxed{a_n\ =\ (-1)^nn(n-1)}.$$

mathmari said:
Ah should it maybe be $$a_{n+1}=(-1)^{n+1}\cdot (|a_n|+2\cdot n)$$ ?
Yes, that should work. (Nod)

Olinguito said:
Thanks, ILS. When the first term is $0$ it’s easy to slip into thinking it’s $a_0$ rather than $a_1$. In this case the formula should be
$$\boxed{a_n\ =\ (-1)^nn(n-1)}.$$

Yes, that should work. (Nod)

So, is the recursive formula \begin{align*}&a_0=0 \\ &a_{n+1}=(-1)^{n+1}\cdot (|a_n|+2\cdot n)\end{align*} and the explicit formula $$a_n\ =\ (-1)^nn(n-1)$$ ? (Wondering)

mathmari said:
So, is the recursive formula \begin{align*}&a_0=0 \\ &a_{n+1}=(-1)^{n+1}\cdot (|a_n|+2\cdot n)\end{align*} and the explicit formula $$a_n\ =\ (-1)^nn(n-1)$$ ?

Yes.
And an alternative form for the recursive formula is $a_{n+1}=- a_n + (-1)^{n+1}\cdot 2n$. (Nerd)

I like Serena said:
Yes.
And an alternative form for the recursive formula is $a_{n+1}=- a_n + (-1)^{n+1}\cdot 2n$. (Nerd)

Why is $(-1)^{n+1}|a_n|=-a_n$ ? (Wondering)

mathmari said:
Why is $(-1)^{n+1}|a_n|=-a_n$ ?

Because $a_n$ alternates in sign.
That is, when $n$ is even, $a_n$ is positive.
And when $n$ is odd, $a_n$ is negative with a special case for $n=1$ since $a_1=0$.

I like Serena said:
Because $a_n$ alternates in sign.
That is, when $n$ is even, $a_n$ is positive.
And when $n$ is odd, $a_n$ is negative with a special case for $n=1$ since $a_1=0$.

I got stuck right now.. Do we have $a_0=0$ or $a_1=0$ ?

I mean do we have \begin{align*}&a_0=0 \\ &a_{n+1}=- a_n + (-1)^{n+1}\cdot 2n\end{align*} or \begin{align*}&a_1=0 \\ &a_{n+1}=- a_n + (-1)^{n+1}\cdot 2n\end{align*} ? (Wondering)

From $$a_{n+1}=- a_n + (-1)^{n+1}\cdot 2n\Rightarrow a_{n+1}+ a_n = (-1)^{n+1}\cdot 2n$$ we get the following equations $$a_{n+1}+ a_n = (-1)^{n+1}\cdot 2n \\ a_{n}+ a_{n-1} = (-1)^{n}\cdot 2(n-1) \\ a_{n-1}+ a_{n-2} = (-1)^{n-1}\cdot 2(n-2) \\ \vdots \\ a_{2}+ a_1 = (-1)^{2}\cdot 2\cdot 1$$ so we don't get an telescoping sum, do we? (Wondering)

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mathmari said:
I got stuck right now.. Do we have $a_0=0$ or $a_1=0$ ?

I mean do we have \begin{align*}&a_0=0 \\ &a_{n+1}=- a_n + (-1)^{n+1}\cdot 2n\end{align*} or \begin{align*}&a_1=0 \\ &a_{n+1}=- a_n + (-1)^{n+1}\cdot 2n\end{align*} ?

In post #1 we were given that $a_1=0$ and $a_0$ is presumably undefined.

mathmari said:
From $$a_{n+1}=- a_n + (-1)^{n+1}\cdot 2n\Rightarrow a_{n+1}+ a_n = (-1)^{n+1}\cdot 2n$$ we get the following equations $$a_{n+1}+ a_n = (-1)^{n+1}\cdot 2n \\ a_{n}+ a_{n-1} = (-1)^{n}\cdot 2(n-1) \\ a_{n-1}+ a_{n-2} = (-1)^{n-1}\cdot 2(n-2) \\ \vdots \\ a_{2}+ a_1 = (-1)^{2}\cdot 2\cdot 1$$ so we don't get an telescoping sum, do we?

It's still a telescoping sum - after we multiply every other line with $-1$. (Smirk)

I like Serena said:
It's still a telescoping sum - after we multiply every other line with $-1$. (Smirk)

Ah ok! So on the left side we get $a_{n+1}$ but which sum do we get on the right sum? (Wondering)

mathmari said:
Ah ok! So on the left side we get $a_{n+1}$ but which sum do we get on the right sum? (Wondering)

Let's consider 2 cases: $n+1$ is even, and $n+1$ is odd.
And let's start with the first one ($n+1$ is even).
What will those equations looks like then? It should simplify them, shouldn't it? (Wondering)

I like Serena said:
Let's consider 2 cases: $n+1$ is even, and $n+1$ is odd.
And let's start with the first one ($n+1$ is even).
What will those equations looks like then? It should simplify them, shouldn't it? (Wondering)

If $n+1$ is even we get $$2n+2(n-1)+2(n-2)+\ldots +2\cdot 2+2\cdot 1=2\sum_{i=1}^{n}i=2\cdot \frac{n(n+1)}{2}=n(n+1)$$
If $n+1$ is odd we get $$-2n-2(n-1)-2(n-2)-\ldots -2\cdot 2-2\cdot 1=-2\sum_{i=1}^{n}i=-2\cdot \frac{n(n+1)}{2}=-n(n+1)$$

Is everything correct? (Wondering)

mathmari said:
If $n+1$ is even we get $$2n+2(n-1)+2(n-2)+\ldots +2\cdot 2+2\cdot 1=2\sum_{i=1}^{n}i=2\cdot \frac{n(n+1)}{2}=n(n+1)$$
If $n+1$ is odd we get $$-2n-2(n-1)-2(n-2)-\ldots -2\cdot 2-2\cdot 1=-2\sum_{i=1}^{n}i=-2\cdot \frac{n(n+1)}{2}=-n(n+1)$$

Is everything correct?
(Nod)

Thanks a lot! (Sun)

## 1. What is the difference between a recursive formula and an explicit form?

A recursive formula is a mathematical equation that defines a sequence by using previous terms in the sequence. An explicit form, on the other hand, directly calculates the value of a term in the sequence without using previous terms.

## 2. Why would someone want to convert a recursive formula into an explicit form?

Converting a recursive formula into an explicit form can make it easier to calculate the value of a term in the sequence without having to go through each previous term. It can also make it easier to find patterns and make predictions about the sequence.

## 3. How do you convert a recursive formula into an explicit form?

To convert a recursive formula into an explicit form, you can use algebraic manipulation to isolate the variable you want to find the value of. You may also need to use substitution to replace previous terms in the sequence with their explicit forms.

## 4. Can all recursive formulas be converted into explicit forms?

No, not all recursive formulas can be converted into explicit forms. Some may have complex patterns or rely on multiple previous terms, making it difficult to find an explicit form.

## 5. Are there any disadvantages to using an explicit form instead of a recursive formula?

One disadvantage is that explicit forms may become more complex and difficult to calculate as the sequence progresses, while recursive formulas may become simpler and more efficient. Additionally, explicit forms may not be as easy to spot or understand in some cases.

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