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# help

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A sequence satisifes a_1 = 3 and a_{n + 1} = a_n/(a_n + 1) for n >= 1.  What is a_{14}?

Dec 17, 2019

#1
+24430
+2

A sequence satisifes $$a_1 = 3$$ and $$a_{n + 1} = \dfrac{a_n}{a_n + 1}$$ for $$n \geq 1$$

What is $$a_{14}$$?

$$\begin{array}{|lrcll|} \hline & \mathbf{a_{n + 1}} &=& \mathbf{\dfrac{a_n}{a_n + 1}} \\\\ & \dfrac{1}{a_{n + 1}} &=& \dfrac{a_n + 1}{a_n} \\\\ & \mathbf{\dfrac{1}{a_{n + 1}}} &=& \mathbf{1+ \dfrac{1}{a_n}} \\\\ \hline \\ & \dfrac{1}{a_{n + 2}} &=& 1+ \dfrac{1}{a_{n+1}} \quad | \quad \mathbf{\dfrac{1}{a_{n + 1}}=1+ \dfrac{1}{a_n}} \\\\ & \dfrac{1}{a_{n + 2}} &=& 1+ 1+ \dfrac{1}{a_n} \\\\ & \mathbf{\dfrac{1}{a_{n + 2}}} &=& \mathbf{2+ \dfrac{1}{a_n}} \\\\ \hline \\ & \dfrac{1}{a_{n + 3}} &=& 1+ \dfrac{1}{a_{n+2}} \quad | \quad \mathbf{\dfrac{1}{a_{n + 2}}=2+ \dfrac{1}{a_n}} \\\\ & \dfrac{1}{a_{n + 3}} &=& 1+ 2+ \dfrac{1}{a_n} \\\\ & \mathbf{\dfrac{1}{a_{n + 3}}} &=& \mathbf{3+ \dfrac{1}{a_n}} \\\\ \hline \\ & \dfrac{1}{a_{n + 4}} &=& 1+ \dfrac{1}{a_{n+3}} \quad | \quad \mathbf{\dfrac{1}{a_{n + 3}}=3+ \dfrac{1}{a_n}} \\\\ & \dfrac{1}{a_{n + 4}} &=& 1+ 3+ \dfrac{1}{a_n} \\\\ & \mathbf{\dfrac{1}{a_{n + 4}}} &=& \mathbf{4+ \dfrac{1}{a_n}} \\ \hline & \ldots \\ \hline \\ & \mathbf{\dfrac{1}{a_{n + m}}} &=& \mathbf{m+ \dfrac{1}{a_n}} \\\\ n=1,\ m=13: & \mathbf{\dfrac{1}{a_{1 + 13}}} &=& \mathbf{13+ \dfrac{1}{a_1}} \\\\ & \dfrac{1}{a_{14}} &=& 13+ \dfrac{1}{a_1} \quad | \quad a_1 =3 \\\\ & \dfrac{1}{a_{14}} &=& 13+ \dfrac{1}{3} \\\\ & \dfrac{1}{a_{14}} &=& \dfrac{3*13+1}{3} \\\\ & \dfrac{1}{a_{14}} &=& \dfrac{40}{3} \\\\ & \mathbf{ a_{14} } &=& \mathbf{\dfrac{3}{40} } \\ \hline \end{array}$$

Dec 17, 2019
#2
+1

The sequence goes like this:
3,  3/4,  3/7,  3/10,  3/13,  3/16,  3/19,.....etc.
The closed form of the sequence is:
a_n = 3/(3 n - 2)
a(14) = 3 / (3*14 - 2)
a(14) = 3 / 40

Dec 17, 2019