Answer to Question #248656 in Discrete Mathematics for Help

Question #248656

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

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


1
Expert's answer
2021-10-11T05:49:49-0400

Let us find a recursive definition of the sequence with general term

tn=6(n+1)!3n,n1.t_n = \frac{6 (n + 1)!}{3^n}, n \ge 1.

It follows that

tn+1=6(n+2)!3n+1=6(n+1)!(n+2)3n3=6(n+1)!3nn+23=tnn+23.t_{n+1} = \frac{6 (n + 2)!}{3^{n+1}}=\frac{6 (n + 1)!(n+2)}{3^{n}\cdot3}=\frac{6 (n + 1)!}{3^{n}}\frac{n+2}{3}=t_n\frac{n+2}{3}.

We conclude that the recursive definition of this sequence is

tn+1=13(n+2)tn, t1=4.t_{n+1}=\frac{1}3(n+2)t_n,\ t_1=4.


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment