There are as many possible answers as there are people answering. Here is one.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?
\(\displaystyleHi 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
\(\displaystyle
a_{k+1}=3^{(k+1)-1} = 3^{(k-1)+1}= 3^{k-1}3^1\space =\space 3\space 3^{k-1}\space=\space ????\)
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: .\(\displaystyle a_n \;=\;a_{n-1} + 2n-1\)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?