Question 1176367
<pre>

This problem certainly does not "need" induction to prove b/c the terms are monotonically increasing and they start with {{{a[1] = 6}}} so all {{{a[n]>5}}} ... but here is the inductive proof:

Base case:
{{{a[1] = 6 }}}  6 > 5, so the base case is true

Hypothesis:
{{{a[n] > 5 }}} for n=k

we also can write
{{{a[k+1] = a[k] + 3k }}}  (*)

Step case:
Let n=k+1.  We must show {{{a[n]>5}}} for n=k+1 (then it holds for all k>=1)
{{{a[(k+1)+1] = a[k+1] + 3(k+1) }}}

Substitute by using (*):
{{{ a[k+2] = (a[k]+3k) + 3(k+1) }}}

Now,  {{{ a[k]+3k }}} > 5 by the hypothesis  (and  3(k+1) > 0  since k>0) 
so the RHS is greater than 5 and {{{a[n] > 5}}} for n=k+1.

Therefore {{{a[n] > 5}}} for {{{k>=1}}} &#9632;


In my opinion, much better introductory problems exist for teaching proof by induction.