Chapter 10.4, Problem 26E

To determine

To write: The first six terms of the sequence.

Answer to Problem 26E

Thefirst six terms of the given $n^{th}$ term are $2,6,12,10,5,3.5$ .

Explanation of Solution

Given:

  $a_1=2,a_n=\frac{(n+1)a_{n-1}}{a_{n-2}}$

Calculation:

The first six terms of the sequence $a_1=2,a_n=\frac{(n+1)a_{n-1}}{a_{n-2}}$ .

The $n^{th}$ term is

  $a_n=\frac{(n+1)a_{n-1}}{a_{n-2}}\mathrm{.....}(1)$

And the first term is

  $a_1=\mathrm{2......}(2)$

The first term is $a_1=\mathrm{2......}(2)$

  $\boxed{a_1=2}$

The second term is

  $\boxed{a_2=6}$

The third term is ,

  $\begin{array}{l}{a}_3=\frac{(3+1)a_{3-1}}{a_{3-2}}\\ a_3=\frac{4(a_2)}{a_1}\\ a_3=\frac{4(6)}{2}\\ \boxed{a_3=12}\end{array}$

The fourth term is,

  $\begin{array}{l}{a}_4=\frac{(4+1)a_{4-1}}{a_{4-2}}\\ a_4=\frac{5(a_3)}{a_2}\\ a_4=\frac{5(12)}{6}\\ \boxed{a_4=10}\end{array}$

The fifth term is ,

  $\begin{array}{l}{a}_5=\frac{(5+1)a_{5-1}}{a_{5-2}}\\ a_5=\frac{6(a_4)}{a_3}\\ a_5=\frac{6(10)}{12}\\ \boxed{a_5=5}\end{array}$

The sixth term is,

  $\begin{array}{l}{a}_6=\frac{(6+1)a_{6-1}}{a_{6-2}}\\ a_6=\frac{7(a_5)}{a_4}\\ a_6=\frac{7(5)}{10}\\ \boxed{a_6=3.5}\end{array}$

Hence, the first six terms of the given $n^{th}$ term are $2,6,12,10,5,3.5$ .

Conclusion:

Therefore, the first six terms of the given $n^{th}$ term are $2,6,12,10,5,3.5$ .