SOLUTION: ) Let the sequence (an) be defined by
an+1 = an + 3n for n ≥ 1 and a1 = 6.
Prove by induction that
an > 5.
Algebra ->
Sequences-and-series
-> SOLUTION: ) Let the sequence (an) be defined by
an+1 = an + 3n for n ≥ 1 and a1 = 6.
Prove by induction that
an > 5.
Log On
This problem certainly does not "need" induction to prove b/c the terms are monotonically increasing and they start with so all ... but here is the inductive proof:
Base case: 6 > 5, so the base case is true
Hypothesis: for n=k
we also can write (*)
Step case:
Let n=k+1. We must show for n=k+1 (then it holds for all k>=1)
Substitute by using (*):
Now, > 5 by the hypothesis (and 3(k+1) > 0 since k>0)
so the RHS is greater than 5 and for n=k+1.
Therefore for ■
In my opinion, much better introductory problems exist for teaching proof by induction.