I see you already did the ones where you figured out the formula. There's a fun way that doesn't require figuring out the formula, called a forward difference table.
Let's work out the first one to learn the technique. We write the sequence and under it the forward differences,
2 5 10 17 26
3 5 7 9
2 2 2
We got 3=5-2, 5=10-5, 7=17-10, etc and the same for subsequent lines. Eventually we get to an all constant line and we stop.
To extend the table we just add some constants to the constant line and work our way back up:
2 5 10 17 26 37 50
3 5 7 9 11 13
2 2 2 2 2
So we get 11 as 9+2 and 37 as 26+11 for our answer which matches the answer you already got.
Let's do the ones you haven't.
2 8 18 32 50 72
6 10 14 18 22
4 4 4 4
Answer: 72
3 9 19 33 51 73
6 10 14 18 22
4 4 4 4
Answer: 73
2 11 26 47 74 107
9 15 21 27 33
6 6 6 6
Answer: 107
Three's our limit so I'll leave the last one to you.
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A comment says we need the functions. They're all quadratics. I'll tell you how they want you to do it and how to do it in general.
They just want you to recognize these are all pretty close to just . The one that goes 3 12 27 etc. we recognize as just
. The one that goes 2 8 18 is just
. The one that goes 3 9 19 is one more than that so
. The one that goes 2 11 26... is one less than 3 12 27 so is
. Again I'll leave the last one.
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We can do these quadratics in general. Let's work from our difference table. If we have
f(n) = an^2 + bn + c
and we form the difference table for f(1), f(2), f(3) we get
a+b+c 4a+2b+c 9a+3b+c
3a+b 5a+b
2a
so we can use the first columns of our difference tables to get our polynomial.
2 8 18 32 50 72
6 10 14 18 22
4 4 4 4
Answer: 72
So we have 2a=4, so a=2
3a+b=6, so b=6-3(2)=0
a+b+c=2 so c=2-0-2=0
That gives us the obvious answer
3 9 19 33 51 73
6 10 14 18 22
4 4 4 4
Answer: 73
We have 2a=4 again so a=2, 3a+b=2 so b=0, and a+b+c=3 so c=1 giving
2 11 26 47 74 107
9 15 21 27 33
6 6 6 6
Answer: 107
2a=6 gives a=3
3a+b=9 gives b=0
a+b+c=2 gives c= -1
