Math question for homework... really hard?

Question

I have to write a formula for the following pattern:
1,5,12,22,35 etc.
(the difference between the numbers goes up by three.. eg. 5-1=4, 12-5=7, 22-12=10 etc.)
I can use as many variables as needed, and I have to find a formula that applies for any given whole number...

Answer 1

an= A·n^2+B·n+C and

1=A+B+C
5=4A+2B+C --> 4=3A+B
12=9A+3B+C --> 7=5A+B --> 3=2A

--> A=3/2 ; B=-1/2 ; C=0

--> an=3/2·n^2-1/2·n = (3n-1)·n / 2

Comp. a1=2·1/2=1 and a4=11·4/2=22 Yes!

Saludos.

Answer 2

Here's a general principle: The differences
between the numbers form an arithmetic progression,
so the formula you are seeking is quadratic.
So let's assume
f(n) = an² + bn + c
and see what we come up with.
We need to plug in 3 values of n and solve for a, b and c.
f(1) = 1 = a+b+c
f(2) = 5 = 4a + 2b + c
f(3) = 12 = 9a + 3b + c.
So, let's solve this system:
Subtract 1st equation from 2nd and 2nd from 3rd to get
3a + b = 4
5a + b = 7
So 2a = 3
a = 3/2
9/2 + b = 8/2
b = -1/2
Finally, a+b+c = 1
so c = 0
The formula is therefore
f(n) = n(3n-1)/2 .
The numbers in your original pattern are
called pentagonal numbers.
Consult
http://en.wikipedia.org/wiki/Pentagonal_number
for more info on these numbers and a
fascinating diagram involving pentagons!