Is there a general procedure for inducing a formula / recognizing a pattern? For some problems, it's pretty simple to deduce what a formula should look like after you enumerate a few examples, but for some problems, it's not so clear. For these less clear examples, is there some procedure I can follow to find the formula?
For example, just now, I was working on the problem:

Find the expected number of tosses it takes to get $k$ 6s in a row.

The recursive formula is
$$
E[N_k ] = 6(E[N_{k - 1}] + 1)
$$
where $N_k$ is the number of tosses it takes to get $k$ 6s in a row. The initial condition is $E[N_1] = 6$ because $N_1$ is a geometric random variable with $1/6$ probability of success.
Then I enumerated a few cases:
$$
E[N_1] = 6 \\
E[N_2] = 42 \\
E[N_3] = 258 \\
E[N_4] = 1554
$$
It's not obvious to me what the formula should be. But I think it should look something like
$$
E[N_k] = 6^k + (k - 1) * \text{something} + \text{maybe other terms}
$$
But I don't know what this "something" and "maybe other terms" should be. If I sit here long enough I could probably figure it out, but is there some kind of general procedure that I can apply to a wide array of problems?

If you call $a_k=E[N_k]$ then you end up with a linear induction relation: $$a_k-6a_{k-1}=6$$
Since it is linear, it solves as the sum of general solution of homegenous equation $H_k-6H_{k-1}=0$, which is $H_k=6^kH_0$ and one particular solution of the full equation.
The theory says that if the RHS is of the form $P(k)r^k$ then you can find a particular solution of the form $Q(k)r^k$ with $\deg(Q)=\deg(P)+m$ where $m$ is the multiplicity of the root of the homogenous equation.
In this case $RHS = 6\times 1^k$ and the single root of homogenous equation is $r=6$. So the multiplicity of $1$ is just $m=0$ (i.e. not a root) this means $\deg(Q)=0+0=0$.
All that to say that our particular solution is simply a constant...
Therefore let search for $S_k=c$ then $c-6c=6\iff c=-\frac 65$
Our general solution is then $$a_k=H_k+S_k=6^kH_0-\frac 65$$
We now solve for initial conditions:
$a_1=6=6H_0-\frac 65\iff H_0=\frac 65$ and we get $$E[N_k]=\frac 65(6^k-1)$$
