recurrence relation

[markraz]

Can someone help me with this. sequence 1 4 9 16 25


n=index

n= 1 2 3 4 5
1 4 9 16 25


Explicit Formula I think is N^2
what is the method to come up with a recurrence relation?

thanks in advance

 

[pka]

Can someone help me with this. sequence 1 4 9 16 25
n=index
n= 1 2 3 4 5
1 4 9 16 25
Explicit Formula I think is N^2
what is the method to come up with a recurrence relation?

There are as many possible answers as there are people answering. Here is one.

\(\displaystyle (n+1)^2 =n^2+2n+1=n(n+2)+1\)

 

[HallsofIvy]

That is not a true "recurrence relation" because, taking \(\displaystyle a_n= n^2\), it does not relate \(\displaystyle a_{n+1}\) to \(\displaystyle a_n\). One way to do this is to note that \(\displaystyle (n+1)^2- n^2= n^2+ 2n+ 1- n^2= 2n+ 1\) so that we can write \(\displaystyle (n+1)^2= n^2+ 2n+ 1\) so that \(\displaystyle a_{n+1}= a_n+ 2n+ 1\). A better answer, I think, would be to eliminate that "w=2n" in terms of \(\displaystyle a_n\).

 

[Ishuda]

Generally, in trying to find a recurrence relationship one writes down the (n+1)^st term and sees what needs to added (to include subtracted/multiplied/divided/...) to the n^th term to change it into the (n+1)^st term. For example, suppose we had the sequence
1, 2, 3, 4, ...
then
a_n = n
a_n+1 = n + 1 = a_n + 1
and we have the recurrence relationship
a_n+1 = a_n + 1

 

[markraz]

another recurrence relation question

Hi sorry for another RR post, I have this other sequence

1 3 9 27 81 243....... start index is k=1


For Explicit formula I get: \(\displaystyle 3^{k-1}\)

For recurrence relation I get: \(\displaystyle 3a_{k}\)

Question.... How do you get from the "Explicit formula" to the "Recurrence Relation" mathematically/systematically?

So far I set it up as follows:

\(\displaystyle a_{k}=3^{k-1}
\)

\(\displaystyle
a_{k+1}=3^{(k+1)-1}
\)

Anyone know what's next? to get to Recurrence Relation: \(\displaystyle 3a_{k}\) ??

thanks

 

[Ishuda]

Hi sorry for another RR post, I have this other sequence

1 3 9 27 81 243....... start index is k=1


For Explicit formula I get: \(\displaystyle 3^{k-1}\)

For recurrence relation I get: \(\displaystyle 3a_{k}\)

Question.... How do you get from the "Explicit formula" to the "Recurrence Relation" mathematically/systematically?

So far I set it up as follows:

\(\displaystyle a_{k}=3^{k-1}
\)

\(\displaystyle
a_{k+1}=3^{(k+1)-1}
\)

Anyone know what's next? to get to Recurrence Relation: \(\displaystyle 3a_{k}\) ??

thanks

\(\displaystyle
a_{k+1}=3^{(k+1)-1} = 3^{(k-1)+1}= 3^{k-1}3^1\space =\space 3\space 3^{k-1}\space=\space ????\)

 

[markraz]

\(\displaystyle
a_{k+1}=3^{(k+1)-1} = 3^{(k-1)+1}= 3^{k-1}3^1\space =\space 3\space 3^{k-1}\space=\space ????\)

ooohh... So you are "distributing" the 3??

\(\displaystyle
3^{(k-1)+1}= 3^{k-1}3^1\) ??


thanks!

 

[soroban]

Hello, markraz!

Can someone help me with this sequence?. . \(\displaystyle \begin{array}{c|ccccc} n & 1& 2& 3& 4& 5 \\ \hline a_n & 1 &4 &9 &16 &25\end{array}\)Explicit formula I think is \(\displaystyle n^2.\) Yes!What is the method to come up with a recurrence relation?

Note each term and its relationship to its preceding term.. . \(\displaystyle \begin{array}{ccccc}a_1 &=& 1 \\ a_2 &=& 1 \color{red}{+ 3} &= &4\\ a_3 &=& 4 \color{red}{+ 5} &=& 9 \\ a_4 &=& 9\color{red}{+7} &=& 16 \\ a_5 &=&16 \color{red}{+ 9} &=& 25 \end{array}\)The \(\displaystyle n^{th}\) term is the preceding term plus an odd number,. . namely, \(\displaystyle 2n-1.\)Therefore: .\(\displaystyle a_n \;=\;a_{n-1} + 2n-1\)