What is the rule for the sequence 3,5,8,13,21,…?
Answer 1:
nth term plus the nth + 1 term:
Explanation:
This sequence is the:
nth term plus the nth + 1 term:
3 + 5 = 8, 5 + 8 = 13, 8 + 13 = 21, 13 + 21 = 34
This is also called the Fibonacci Series.
Answer 2:
The general term is given by the formula:
\(a_{n}=\left(\frac{3}{2}+\frac{7}{10} \sqrt{5}\right)\left(\frac{1}{2}+\frac{\sqrt{5}}{2}\right)^{n-1}+\left(\frac{3}{2}-\frac{7}{10} \sqrt{5}\right)\left(\frac{1}{2}-\frac{\sqrt{5}}{2}\right)^{n-1}\).
Explanation:
The Fibonacci sequence is defined by:
\(F_{0}=0\)
\(F_{1}=1\)
\(F_{n+2}=F_{n}+F_{n+1}\)
The first few terms are:
0,1,1,2,3,5,8,13,21,34,55,89,144,…
Note that the given sequence starts at \(F_{4}=3\), but otherwise follows the same rules.
\(a_{1}=3\)
\(a_{2}=5\)
\(a_{n+2}=a_{n}+a_{n+1}\)
In order to find a general formula consider the geometric sequence:
\(1, x, x^{2}, \ldots\)If this sequence satisfies the same recursive rule as the Fibonacci sequence then:
\(x^{2}=1+x\)So:
where \(A\) and \(B\) are any constants.
Notice that any such sequence \(b_{n}\) will satisfy the recursive rule:
So if we can find values for \(A\) and \(B\) such that\(b_{1}=a_{1}\) and \(b_{2}=a_{2}\), then we have a general formula for our sequence.
So we just require:
Subtracting \(\frac{3}{2}\) from both ends of this second equation, we get:
Multiply both sides by \(\frac{2}{\sqrt{5}}\) to get:
\(\frac{7}{\sqrt{5}}=A-B\)Adding this to the first equation we find:
\(2 A=3+\frac{7}{\sqrt{5}}=3+\frac{7}{5} \sqrt{5} \quad \text { so } \quad A=\frac{3}{2}+\frac{7}{10} \sqrt{5}\)Subtracting from the first equation we find:
\(2 B=3-\frac{7}{\sqrt{5}}=3-\frac{7}{5} \sqrt{5} \quad \text { so } \quad B=\frac{3}{2}-\frac{7}{10} \sqrt{5}\)Hence the general formula for the given sequence can be written:
\(a_{n}=\left(\frac{3}{2}+\frac{7}{10} \sqrt{5}\right)\left(\frac{1}{2}+\frac{\sqrt{5}}{2}\right)^{n-1}+\left(\frac{3}{2}-\frac{7}{10} \sqrt{5}\right)\left(\frac{1}{2}-\frac{\sqrt{5}}{2}\right)^{n-1}\)