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^+#.
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