+0

much help appreciated!

0
185
3

Let $$a_0=6$$ and $$a_n = \frac{a_{n - 1}}{1 + a_{n - 1}}$$ for all $$n \ge 1$$. Find $$a_{100}$$

Mar 17, 2020

#1
+21959
+1

a =  6          an  =  an-1 / ( 1 + an-1 )

a1  =  6 / ( 1 + 6 )  =  6/7

a2  =  (6/7) / ( 1 + 6/7 )  =  (6/7) / (13/7)  =  6/13

a3  =  (6/13) / ( 1 + 6/13 )  =  (6/13) / (19/13)  =  6/19

Notice that the numerator is always 6 and the denominator is 6 times the term's subscript + 1.

So for the fourth term, the numerator will be 6 and the denominator will be 6 x 4 + 1  =  25.  a4  =  6/25.

Use this to find a100.

Mar 17, 2020
#2
+1

thank you!!

Mar 17, 2020
#3
+25656
+2

Let $$a_0=6$$ and $$a_n = \dfrac{a_{n - 1}}{1 + a_{n - 1}}$$ for all $$n \ge 1$$.
Find $$a_{100}$$

$$\begin{array}{|rcll|} \hline a_n &=& \dfrac{a_{n - 1}}{1 + a_{n - 1}} \\\\ \dfrac{1}{a_n} &=& \dfrac{1 + a_{n - 1}} {a_{n - 1}} \\\\ \dfrac{1}{a_n} &=& \dfrac{1} {a_{n - 1}} +\dfrac{a_{n - 1}} {a_{n - 1}} \\\\ \mathbf{\dfrac{1}{a_n}} &=& \mathbf{\dfrac{1} {a_{n - 1}} + 1} \\ \hline \end{array}$$

$$\begin{array}{|lrcll|} \hline \mathbf{\dfrac{1}{a_n}} = \mathbf{\dfrac{1} {a_{n - 1}} + 1} \\ \hline n=1: & \dfrac{1}{a_1} &=& \dfrac{1} {a_{1 - 1}} + 1 \\\\ & \mathbf{\dfrac{1}{a_1}} &=& \mathbf{ \dfrac{1} {a_{0}} + 1} \\ \hline n=2: & \dfrac{1}{a_2} &=& \dfrac{1} {a_{2-1}} + 1 \\\\ & \dfrac{1}{a_2} &=& \dfrac{1} {a_{1}} + 1 \\\\ & \dfrac{1}{a_2} &=& \dfrac{1} {a_{0}} + 1 + 1 \\\\ & \mathbf{\dfrac{1}{a_2}} &=& \mathbf{\dfrac{1} {a_{0}} + 2} \\ \hline n=3: & \dfrac{1}{a_3} &=& \dfrac{1} {a_{3-1}} + 1 \\\\ & \dfrac{1}{a_3} &=& \dfrac{1} {a_{2}} + 1 \\\\ & \dfrac{1}{a_3} &=& \dfrac{1} {a_{0}} + 2 + 1 \\\\ & \mathbf{\dfrac{1}{a_3}} &=& \mathbf{\dfrac{1} {a_{0}} + 3} \\ \hline \ldots \\ \hline & \mathbf{\dfrac{1}{a_n}} &=& \mathbf{\dfrac{1} {a_{0}} + n} \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline \mathbf{\dfrac{1}{a_n}} &=& \mathbf{\dfrac{1} {a_{0}} + n} \quad | \quad a_0 = 6 \\\\ \dfrac{1}{a_n} &=& \dfrac{1} {6} + n \quad | \quad n = 100 \\\\ \dfrac{1}{a_{100}} &=& \dfrac{1} {6} + 100 \\\\ \dfrac{1}{a_{100}} &=& \dfrac{1} {6} + \dfrac{600}{6} \\\\ \dfrac{1}{a_{100}} &=& \dfrac{1+600} {6} \\\\ \dfrac{1}{a_{100}} &=& \dfrac{601} {6} \\\\ \mathbf{ a_{100} } &=& \mathbf{\dfrac{6}{601} } \\ \hline \end{array}$$

Mar 17, 2020