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# Sequence problem

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The sequence $$\{a_n\}$$ is defined by

$$a_0 = 1,a_1 = 1, \text{ and } a_n = a_{n - 1} + \frac {a_{n - 1}^2}{a_{n - 2}}\text{ for }n\ge2.$$

The sequence $$\{b_n\}$$ is defined by

$$b_0 = 1,b_1 = 3, \text{ and } b_n = b_{n - 1} + \frac {b_{n - 1}^2}{b_{n - 2}}\text{ for }n\ge2.$$

Find $$\frac {b_{32}}{a_{32}}.$$

Aug 11, 2019

#1
+1

a - 1, 1, 2, 6, 24, 120, 720, 5040, 40320, 362880, 3628800, 39916800, 479001600, ...
a(n) =1(n - 1) .......This is called "rising factorial"

b - 1, 3, 12, 60, 360, 2520, 20160, 20160, 181440, 1814400, 19958400, 239500800, 3113510400, ...
a(n) =1(n + 1) / 2 ............This is called "rising factorial"

b(32) / a(32) =[1(32+1)! / 2] / 1[32 - 1]! =[33!/2] / 31! = 528

Aug 12, 2019
#2
+24389
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Sequence problem
The sequence $$\{a_n\}$$ is defined by
$$a_0 = 1,a_1 = 1, \text{ and } a_n = a_{n - 1} + \dfrac {a_{n - 1}^2}{a_{n - 2}}\text{ for }n\ge2$$.

The sequence \{b_n\} is defined by
$$b_0 = 1,b_1 = 3, \text{ and } b_n = b_{n - 1} + \dfrac {b_{n - 1}^2}{b_{n - 2}}\text{ for }n\ge2$$.

Find $$\dfrac {b_{32}}{a_{32}}$$.

$$\begin{array}{|rcll|} \hline a_n &=& a_{n - 1} + \dfrac { a_{n - 1}^2 } { a_{n - 2} } \\ \mathbf{ a_n } &=& \mathbf{ a_{n - 1} \left( 1 + \dfrac{a_{n-1}}{a_{n-2}} \right) } \\ \hline a_0 &=& 1 \text{ or } a_0 = 0! \\ a_1 &=& 1 \text{ or } a_1 = 1! \\ \hline a_2 &=& 1!\left(1+\dfrac{1}{1}\right) \\ &=& 1!\cdot 2 \\ &=& \mathbf{2!} \\\\ a_3 &=& 2!\left(1+\dfrac{1!\cdot 2}{1!}\right) \\ &=& 2!\cdot 3 \\ &=& \mathbf{3!} \\\\ a_4 &=& 3!\left(1+\dfrac{2!\cdot 3}{2!}\right) \\ &=& 3!\cdot 4 \\ &=& \mathbf{4!} \\\\ a_5 &=& 4!\left(1+\dfrac{3!\cdot 4}{3!}\right) \\ &=& 4!\cdot 5 \\ &=& \mathbf{5!} \\ \ldots \\ \mathbf{a_n} &=& \mathbf{n!} \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline b_n &=& b_{n-1} + \dfrac{ b_{n-1}^2 } { b_{n-2} } \\ \mathbf{ b_n } &=& \mathbf{ b_{n-1} \left( 1 + \dfrac{b_{n-1}}{b_{n-2}} \right) } \\ \hline b_0 &=& 1 \text{ or } b_0 = \dfrac{2!}{2!} \\ b_1 &=& 3 \text{ or } b_1 = \dfrac{3!}{2!} \\ \hline b_2 &=& 3\left(1+\dfrac{3}{1}\right) \\ &=& 3\cdot 4 \\\\ b_3 &=& 3\cdot 4\left(1+\dfrac{\not{3}\cdot 4}{\not{3}}\right) \\ &=& 3\cdot 4\cdot 5 \\\\ b_4 &=& 3\cdot 4\cdot 5\left(1+\dfrac{\not{3}\cdot \not{4}\cdot 5}{\not{3}\cdot\not{4}}\right) \\ &=& 3\cdot 4\cdot 5\cdot 6 \\\\ b_5 &=& 3\cdot 4\cdot 5\cdot 6\left(1+\dfrac{\not{3}\cdot \not{4}\cdot \not{5}\cdot 6}{\not{3}\cdot\not{4}\cdot \not{5}}\right) \\ &=& 3\cdot 4\cdot 5\cdot 6\cdot 7 \\ \ldots \\ \mathbf{b_n} &=& \mathbf{\dfrac{(n+2)!}{2!} } \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline \dfrac{b_n}{a_n} &=& \dfrac{\dfrac{(n+2)!}{2!}} {n!} \\\\ &=& \dfrac{ (n+2)! } {2!n!} \\\\ &=& \dfrac{ n!(n+1)(n+2) } {2!n!} \\\\ &=& \dfrac{ (n+1)(n+2) } {2!} \\\\ \mathbf{ \dfrac{b_n}{a_n} } &=& \mathbf{ \dfrac{(n+1)(n+2)} {2}} \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline \dfrac{b_{32}}{a_{32}} &=& \dfrac{(32+1)(32+2)} {2} \\\\ &=& \dfrac{33\cdot 34}{2} \\\\ &=& 33\cdot 17 \\\\ \mathbf{ \dfrac{b_{32}} {a_{32}} } &=&\mathbf{561} \\ \hline \end{array}$$

Aug 12, 2019