# Can you use mathematical induction to prove that the sequence defined by t_1=6 , t_(n+1)= t_n/(3n for all n in ZZ^+ can be written as t_n=18/(3^n(n-1)! for all n in ZZ^+?

Feb 4, 2017

Proof: (By induction)

Base case: For $n = 1$ we have t_1 = 6 = 18/(3^1*0!)

Inductive hypothesis: Suppose that t_k = 18/(3^k(k-1)!) for some $k \in {\mathbb{Z}}^{+}$.

Induction step: We wish to show that t_(k+1) = 18/(3^(k+1)*k!). Indeed,

${t}_{k + 1} = \frac{{t}_{k}}{3 k}$

=(18/(3^k(k-1)!))/(3k)

=18/(3*3^k*k*(k-1)!)

=18/(3^(k+1)*k!)

as desired.

We have supposed true for $k$ and shown true for $k + 1$, thus, by induction, t_n = 18/(3^n(n-1)!), AAn in ZZ^+.