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 $n in ZZ^+$ ?

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Answer 1 · 2017-02-04T10:30:52

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 $k in ZZ^+$ .

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

$t_(k+1) = (t_k)/(3k)$

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

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

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

as desired.

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

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