SOLUTION: What is the nth term for the sequence 3,12,27,48,75....

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Question 1100724: What is the nth term for the sequence 3,12,27,48,75....
Found 3 solutions by richwmiller, ikleyn, greenestamps:
Answer by richwmiller(17219) About Me  (Show Source):
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See # Question 1100724 Answer by MathLover1
take the differences between the numbers to get:
9, 15, 21, 27
take the second differences by taking the differences of the numbers above:
6, 6, 6
all the second differences are equal to six, which means the equation is quadratic: y=ax^2+bx+c
we can find the a, b, & c values by pluggin in the points (1,3), (2,12), & (3,27) and solving for a,b,& c by elimination:
1a+1b+1c=3
4a+2b+1c=12
9a+3b+1c=27
1a+1b+1c=3
3a+1b+0c=9
8a+2b+0c=24

1a+1b+1c=3
3a+1b+0c=9
2a+0b+0c=6
2a=6
a=3
3a+b=9
b=0
a+b+c=3
c=0
final equation of nth term is: 3n%5E2%2B0n%2B0+=+3n%5E2 or
+3n%5E2

Answer by ikleyn(52788) About Me  (Show Source):
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.
The sequence is 

3 = a%5B1%5D = 3,

12 = a%5B2%5D = 3%2A2%5E2,

27 = a%5B3%5D = 3%2A3%5E2,

48 = a%5B4%5D = 3%2A4%5E2,

75 = a%5B5%5D = 3%2A5%5E2

. . . . 

n-th term = a%5Bn%5D = 3%2An%5E2.

I don't like very much the problems of the type "guess the next number".

I think that they all are a mix of Alchemy and Astrology and have nothing common with the Math.

But in this case the pattern is obvious . . .

I assume that other teachers (and I even know who) will come to argue my solution and offer theirs . . .

I have nothing against of it.

It will only confirm that NEITHER questions NOR answers make sense in this area . . .



Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


Just to add to what the other two tutors said about this....

Tutor ikleyn is exactly right, in that NO problem of this type has a single correct answer. Given any finite sequence of numbers, you can put any numbers after the given numbers and it will be a valid sequence. And in general the resulting sequence will not have a formula for the n-th term.

However, if it is a worthwhile problem, there should be a relatively simple pattern to the sequence, and hence a formula for the n-th term.

Sometimes you can spot the pattern easily, as many people might be able to with this example.

If there is not easily identifiable pattern, the problem is nearly always not worth the time and effort to try to solve, because you will only be guessing what pattern the author of the problem used.

Finally, given any finite sequence of numbers, you can find a polynomial that produces that sequence, using the method of finite differences that the first tutor used.

If you are interested in learning more about that method, just do an internet search on "finite differences".

But only serious math nerds (like me!) will go to that kind of trouble.

The best thing to do with these kinds of problems is spend a minute or two looking for a simple pattern and then find something better to do with your time.