Let \(a_0=6\) and \(a_n = \frac{a_{n - 1}}{1 + a_{n - 1}}\) for all \(n \ge 1\). Find \(a_{100}\)
+23245 a0 = 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.
+26367 Let \(a_0=6\) and \(a_n = \dfrac{a_{n - 1}}{1 + a_{n - 1}}\) for all \(n \ge 1\).
Find \(a_{100}\)
\(\begin{array}{|rcll|} \hline a_n &=& \dfrac{a_{n - 1}}{1 + a_{n - 1}} \\\\ \dfrac{1}{a_n} &=& \dfrac{1 + a_{n - 1}} {a_{n - 1}} \\\\ \dfrac{1}{a_n} &=& \dfrac{1} {a_{n - 1}} +\dfrac{a_{n - 1}} {a_{n - 1}} \\\\ \mathbf{\dfrac{1}{a_n}} &=& \mathbf{\dfrac{1} {a_{n - 1}} + 1} \\ \hline \end{array}\)
\(\begin{array}{|lrcll|} \hline \mathbf{\dfrac{1}{a_n}} = \mathbf{\dfrac{1} {a_{n - 1}} + 1} \\ \hline n=1: & \dfrac{1}{a_1} &=& \dfrac{1} {a_{1 - 1}} + 1 \\\\ & \mathbf{\dfrac{1}{a_1}} &=& \mathbf{ \dfrac{1} {a_{0}} + 1} \\ \hline n=2: & \dfrac{1}{a_2} &=& \dfrac{1} {a_{2-1}} + 1 \\\\ & \dfrac{1}{a_2} &=& \dfrac{1} {a_{1}} + 1 \\\\ & \dfrac{1}{a_2} &=& \dfrac{1} {a_{0}} + 1 + 1 \\\\ & \mathbf{\dfrac{1}{a_2}} &=& \mathbf{\dfrac{1} {a_{0}} + 2} \\ \hline n=3: & \dfrac{1}{a_3} &=& \dfrac{1} {a_{3-1}} + 1 \\\\ & \dfrac{1}{a_3} &=& \dfrac{1} {a_{2}} + 1 \\\\ & \dfrac{1}{a_3} &=& \dfrac{1} {a_{0}} + 2 + 1 \\\\ & \mathbf{\dfrac{1}{a_3}} &=& \mathbf{\dfrac{1} {a_{0}} + 3} \\ \hline \ldots \\ \hline & \mathbf{\dfrac{1}{a_n}} &=& \mathbf{\dfrac{1} {a_{0}} + n} \\ \hline \end{array}\)
\(\begin{array}{|rcll|} \hline \mathbf{\dfrac{1}{a_n}} &=& \mathbf{\dfrac{1} {a_{0}} + n} \quad | \quad a_0 = 6 \\\\ \dfrac{1}{a_n} &=& \dfrac{1} {6} + n \quad | \quad n = 100 \\\\ \dfrac{1}{a_{100}} &=& \dfrac{1} {6} + 100 \\\\ \dfrac{1}{a_{100}} &=& \dfrac{1} {6} + \dfrac{600}{6} \\\\ \dfrac{1}{a_{100}} &=& \dfrac{1+600} {6} \\\\ \dfrac{1}{a_{100}} &=& \dfrac{601} {6} \\\\ \mathbf{ a_{100} } &=& \mathbf{\dfrac{6}{601} } \\ \hline \end{array}\)
