My approach to solving number series, answers for the previous puzzles and two extra challenges

When I'm confronted with a numeric-sequence-style puzzle, one of three things usually happens:

1. I recognize the pattern immediately
consider the following sequence:
1,8,27,64,125,___
I KNOW these are exactly the cubes because I've encountered them enough times throughout my life to recognize them. And so, 6^3=216 is the next term in our series.

2. I recognize the pattern immediately after a small adjustment
now, let's have a look at this example:
-1,2,7,14,23,34,__
In this case, I probably would have noticed that 23 is suspiciously close to 25 and that 34 is suspiciously close to 36. Why? because both 25 and 36 are square numbers. Another closer look would reveal that, in fact, 14 is as "far" from a square number (namely, 2 from 16) as are 23 from 25 and 34 from 36 respectively. From here, it immediately follows that every number in the above sequence is of the form N^2-2 and that our next term is 7^2-2=47.

3. I try to establish a relation of the form aX+b
again, it is probably best to illustrate this point with an example:
4,11,39,151,599,__
Not as obvious, right?
At this point I would ask myself, "How do you get from 4 to 11?" and would try to apply the relation between the terms. Most obvious is the relation "+7" or, as we described above, 1*X+7 or (1,7) in short.
But: 11+7=18≠39, so (1,7) can't be correct.
What about (2,3) and (3,-1)?
2*4+3=11
3*4-1=11
But unfortunately:
2*11+3=25≠39
3*11-1=32≠39
Yiks! I sure hope I won't have to keep checking billions of options! Let's try one more time, (4,-5) seems to work:
4*4-5=11
and amazingly enough it holds for the rest of our terms!
4*11-5=39
4*39-5=151
4*151-5=599
So the next term should be:
4*599-5=2391
   
Another great (and very natural) option would have been to look at the differences between the terms:
11-4=7,39-11=28,151-39=112,599-151=448
here point number 1 kicks in and I immediately recognize these as multiples of 7, namely:
7=7*1,28=7*4,112=7*16,448=7*64
point 1 kicks in AGAIN. The bold numbers are powers of 4:
4^0=1,4^1=4,4^2=16,4^3=64
All that's left is to look at the next logical step:
4^4=256
7*256=1792
599+1792=2391
and amazingly enough, we got the same answer! 
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You might find it surprising, but in probably 75% of all the numeric series I've ever solved (being quite a few, since it is a life-long hobby of mine) the above three (and halfish) approaches/steps were enough. Only in significantly more difficult cases* or cases in which the rule isn't an arithmetic one**, I find myself applying other methods.

*- The "difficulty" of a solution can sometimes be an "all-or-nothing" situation if it heavily relies on your knowledge of specific numerical progressions. Can you solve this one? :
4,6,3,8,1,12,-1,16,___,___

**- Not every sequence containing numbers is making use of an arithmetic relation between the terms. Can you figure this one out? :
365,12,52,7,___,___

Now we finally get to the solutions of last week's puzzles:

Simple Arithmetic
0,2,8,26,80,___
the relation (3,2) holds for all terms (see possibility 3 above), so:
3*80+2=242

-1,-5,15,-85,415,___
the relation (-5,-10) holds for all terms, so:
-5*415-10=-2085

0,-3,6,33,1086,___
6 and 33 are both suspiciously close to a square number (see possibility 2 above). Trying to apply the relation reveals:
0^2-3=0
(-3)^2-3=6
6^2-3=33
33^2-3=1086
so: 1086^2-3=1179393

-2,25,1024,1062961,___
25 and 1024 are squares (5^2 and 32^2).
A more careful look would reveal that 1062961 is also a square (1031^2).
If we will try to establish a relation between 5, 32 and 1031, we will notice that both 32 and 1031 are suspiciously close to a square number (25=5^2<32<36=6^2 and 1024=32^2<1031) and that both 32 and 1031 are +7 away from the lower square, so we can consider the relation (x+7)^2. And in fact:
(-2+7)^2=25
(25+7)^2=1024
(1024+7)^2=1062961
So the next term should be:
(1062961+7)^2=1129900969024
(I was lazy and used a calculator for this one, even though using the binomial formula recursively isn't too terrible).

Not So Simple Arithmetic

3,6,7,10,20,__,__,__
Here we have a complex relation that changes from term to term.
We can easily recognize that:
3*2=6
6+1=7
7+3=10
10*2=20
So why not continuing the cycle? (*2,+1,+3,*2,...):
20+1=21
21+3=24
24*2=48

4,-1,-2,3,-2,-4,__,__,__
Same idea:
4-5=-1
-1*2=-2
-2+5=3
3-5=-2
-2*2=-4
So the cycle is of the form {-5,*2,+5}:
-4+5=1
1-5=-4
-4*2=-8

11,4,16,14,7,__,__,__
Here we have a trickier cycle {-7,^2,-2}:
7^2=49
49-2=47
47-7=40

14,15,3,27,__,__,__
And an even trickier one {+1,-12,^3}:
27+1=28
28-12=16
16^3=4096

A Lovely Pair
1,3,5,11,21,__,__
The trick here is to notice that we have a relation of the form aX+b between two terms. In this case, 2A+B (A and B being the first and second terms, then the second and third and so on).
So:
2*1+3=5
2*3+5=11
2*5+11=21
2*11+21=43
2*21+43=85

-1,-1,2,-7,23,-76,__,__
Same concept with A-3B.
23-3*(-76)=251
-76-3*251=-829

0,0,1,2,8,73,___
A trickier relation, namely (A+1)^2+B^2:
(8+1)^2+73^2=5410

1,2,4,7,12,28,___
Another tricky one, (A-B)^2+3:
(12-28)^2+3=259

And that's all for today! If you find any mistakes or have a suggestion for an interesting puzzle/riddle please let me know!
I hope you had some fun puzzling along, I hope you learned something and stay tuned, next time we'll have a look at another kind of puzzle.