First of all, the phrase "explanatory" in the question should be replaced with "insightful", as I think you're asking for examples of inductive proofs which give the reader some real understanding of what is going on.
A lot of inductive proofs give insight. Here are a few elementary examples.
If $a$ and $b$ are relatively prime positive integers, we can write $ax + by = 1$ for some integers $x$ and $y$. The proof uses induction on $\max(a,b)$ and the inductive step is a replacement of $a$ and $b$ by $a$ and $b-a$ (if $a < b$). The maximum has decreased and this idea in fact leads to Euclid's algorithm, which is the efficient way to practically find $x$ and $y$.
For every positive integer $n$, $\cos(nx)$ is a polynomial in $\cos x$ (e.g., $\cos(2x) = 2\cos(x)^2 - 1$). The proof uses the addition formula for $\cos x$ and that idea even shows you how to build the polynomial expression if you want to find it explicitly, and that may be comforting to some people.
A polynomial of degree $n$ over a field has at most $n$ roots. The proof shows you how the existence of one root controls the cardinality of the number of possible other roots, using some algebra.
Many theorems about polynomials in several variables proceed by induction on the number of variables, and often watching how you can bootstrap a result in $n$ variables to a result in $n+1$ variables gives some kind of nice understanding (if only the understanding that the key case is one variable, suitably formulated). For example, a polynomial in several variables over an infinite field which is identically 0 as a function is the zero polynomial (all coefficients are 0). This is proved by induction and the base case is the previous example, which is really the only involved step.
Here are some theorems where the inductive proof does not give insight.
The "exchange lemma" from linear algebra, which is used to justify why any two bases of a finite-dimensional vector space have the same size, has never seemed particularly enlightening to me. It is proved (in part) using induction.
Different complex-valued characters of a finite abelian group are linearly independent functions. The proof goes by induction on the number of characters, but I never thought the proof itself really explains the linear independence in an "aha" kind of way. It verifies the truth and then you move on to use it.
Cauchy's forwards-backwards inductive proof of the arithmetic-geometric mean inequality is pretty remarkable (since so few theorems are amenable to a forwards-backwards inductive proof [EDIT: an MO question asking for more examples of Cauchy's method is here]), but all the same it doesn't feel like it offers any useful understanding behind the inequality.
The sum of the first $n$ positive integers is $n(n+1)/2$. You can check the identity by induction but the mystery remains of how such a formula is found.
The sum of the squares of the first $n$ positive integers is $n(n+1)(2n+1)/6$. You can check the identity by induction but again there is no real understanding generated of where the identity comes from.
The sum of the cubes of the first $n$ positive integers is $(n(n+1)/2)^2$. You can check the identity by induction but not only does that approach not really explain the formula, it also doesn't explain the surprise that the formula is the square of the formula for the sum of the first $n$ positive integers.
I think I could give an argument by induction to generate infinitely many similar noninsightful proofs by induction starting with item 4, but such an argument wouldn't really give any insight itself, so I will stop here.
