Answer to Question #248132 in Discrete Mathematics for zid

Question

Find, showing all working, a recursive definition of the sequence with general term

tn = 6 (n + 1)!/3n, n >= 1

Answer 1

$$t_1=\dfrac{6(1+1)!}{3(1)}=\dfrac{2(2)!}{1}=4$$

$$t_n=\dfrac{6(n+1)!}{3n}=\dfrac{2(n+1)!}{n}, n\geq1$$

$$t_{n+1}=\dfrac{2(n+1+1)!}{n+1}=\dfrac{2(n+1)!(n+2)}{n+1}$$

$$=\dfrac{2(n+1)!}{n}(\dfrac{n(n+2)}{n+1})=t_n(\dfrac{n(n+2)}{n+1})$$

$$t_{n+1}=t_n\cdot\dfrac{n(n+2)}{n+1}, n\geq1$$