MHB Convert the recursive formula into the explicit form

AI Thread Summary
The discussion revolves around converting a recursive sequence defined as a_1=0 and a_{n+1}=(-1)^{n+1}(a_n + 2n) into an explicit formula. The explicit form derived is a_n = (-1)^{n+1}n(n-1), which captures the alternating signs and the quadratic nature of the sequence. Participants clarify the initial conditions and the implications of the recursive definition, emphasizing the importance of correctly identifying whether a_0 or a_1 is zero. The conversation also touches on the telescoping nature of the sums involved in deriving the explicit formula. Ultimately, the explicit formula and the recursive definition are affirmed as correct, providing a clear understanding of the sequence's behavior.
mathmari
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Hey! :o

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)
 
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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! :o

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 $a_1=0$.
 
  • #10
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)
 
  • #11
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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  • #12
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.
All posts in this thread follow that definition. (Angel)

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)
 
  • #13
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)
 
  • #14
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)
 
  • #15
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)
 
  • #16
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)
 
  • #17
Thanks a lot! (Sun)
 
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