SOLUTION: ) Let the sequence (an) be defined by an+1 = an + 3n for n ≥ 1 and a1 = 6. Prove by induction that an > 5.

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Question 1176367: ) Let the sequence (an) be defined by
an+1 = an + 3n for n ≥ 1 and a1 = 6.
Prove by induction that
an > 5.

Answer by math_helper(2461) About Me  (Show Source):
You can put this solution on YOUR website!

This problem certainly does not "need" induction to prove b/c the terms are monotonically increasing and they start with a%5B1%5D+=+6 so all a%5Bn%5D%3E5 ... but here is the inductive proof:
Base case:
a%5B1%5D+=+6+ 6 > 5, so the base case is true
Hypothesis:
a%5Bn%5D+%3E+5+ for n=k
we also can write
a%5Bk%2B1%5D+=+a%5Bk%5D+%2B+3k+ (*)
Step case:
Let n=k+1. We must show a%5Bn%5D%3E5 for n=k+1 (then it holds for all k>=1)
a%5B%28k%2B1%29%2B1%5D+=+a%5Bk%2B1%5D+%2B+3%28k%2B1%29+
Substitute by using (*):
+a%5Bk%2B2%5D+=+%28a%5Bk%5D%2B3k%29+%2B+3%28k%2B1%29+
Now, +a%5Bk%5D%2B3k+ > 5 by the hypothesis (and 3(k+1) > 0 since k>0)
so the RHS is greater than 5 and a%5Bn%5D+%3E+5 for n=k+1.
Therefore a%5Bn%5D+%3E+5 for k%3E=1

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