TITLE: Proving with Induction When The Claim Fails For Several Numbers
QUESTION [5 upvotes]: I want to find for what positive integers n, the statement $11n+17 \leq 2^n$ is true.
I then want to prove that this is true with induction. The problem I see here, is that to prove it, I need to find P(1), which gives $11+17\leq2$, which is false.

REPLY [5 votes]: I think that in this problem it is instructive to do the induction step first.  Now admittedly this is not following the most standard recipe for induction proofs, and we certainly need to analyze base cases at some point, but doing it this way will answer an important implicit question: how do we know that there is some $N$ such that for all $n \geq N$ we have $11n + 17 \leq 2^n$?  (Well, we should know this from calculus, but how do we see it in this framework?)
For a positive integer $N$, let $P(N)$ be the assertion $11N + 17 \leq 2^N$.
Step 1 (Induction Step): Suppose that $P(N)$ holds for some $N$.  Now we try to prove $P(N+1): 11(N+1) + 17 \leq 2^{N+1}$.  
Note $2^{N+1} = 2 \cdot 2^N = 2^N + 2^N \geq 2^N + 11N + 17$;
the last inequality used our induction hypothesis $P(N)$.  Since $11(N+1) + 17 = (11N+17) + 11$, we will have $2^N + 11N + 17 \geq 11(N+1) + 17$ as long as $2^N \geq 11$.  In turn, this inequality is true for all $N \geq 4$, so if $N \geq 4$ then 
$2^{N+1} \geq 2^N + 2^N \geq 11 + (11N + 17) = 11(N+1) + 17$.
What have we shown?  This:
If if for some $N_0 \geq 4$ we have $11N_0 + 17 \leq 2^{N_0}$, then for all $n \geq N_0$ we have $11n + 17 \leq 2^n$.
Step 2 (Base Cases): checking small positive integers $n$ one finds that $11n + 7 > 2^n$ for $1 \leq n \leq 6$ but $11 \cdot 7 + 7 = 84 < 128 = 2^7$.
Conclusion: For a positive integer $n$, $11n + 7 \leq 2^n$ iff $n \geq 7$.
A comment: what I didn't fully like about some of the other answers was the part where they said: keep trying values of $n$ until you find one for which the inequality holds, then prove by induction that it holds for all greater values of $n$.  This happens to holds here, but it need not hold for other, similar problems.  In fact, compare this to Proposition 6 in these notes (on induction, from a sophomore level class on proof techniques) in which one is looking at the inequality $2^n \geq n^3$ on the natural numbers.  This holds for $n = 0,1$, then fails for $n = 2$ through $9$, then holds again for all integers $n \geq 10$.  On the other hand, the analysis using the induction step shows that if the inequality holds for any $n \geq 4$ then it holds for all larger values of $n$.  In this problem as well it would be more efficient to do the induction step first: otherwise one would have to look at several more individual cases to build up confidence that one has the correct minimal value $N_0$ such that the inequality holds for all $n \geq N_0$.  (I confess though that in my writeup of this problem I do not do the induction step first.  I wanted to stick to the standard recipe as much as possible.)