| |
3 |
|
14 |
|
39 |
|
84 |
|
155 |
|
258 |
| 1st diff |
|
11 |
|
25 |
|
45 |
|
71 |
|
103 |
|
| 2nd diff |
|
|
14 |
|
20 |
|
26 |
|
32 |
|
|
| 3rd diff |
|
|
|
6 |
|
6 |
|
6 |
|
|
|
The table explains how 258 is the next number in the series. The bold underlined numbers are projections of the pattern to show how the next term is calculated. The third difference row shows 6 as the constant between the second differences. This is only one of perhaps several possible solutions.
Alternatively, you might have observed that each term is a multiple of its position in the series. If we divide each term by its position we get: 3,7,13,21,31. The differences between each term: 4,6,8,10. So the next difference would be 12. This gives us 3,7,13,21,31,43. Multiply 43 by 6 to get 258.
Another way of solving it is to work out f(x) where f(x)=ax^3+bx^2+cx+d.
We start with x=0. f(0)=3, first term,=d when x=0, so d=3.
f(1)=14=a+b+c+d=a+b+c+3, so a+b+c=11. Call this equation A.
f(2)=39=8a+4b+2c+3, so 8a+4b+2c=36, eqn B: 4a+2b+c=18.
f(3)=84=27a+9b+3c+3, so 27a+9b+3c=81, eqn C: 9a+3b+c=27.
f(4)=155=64a+16b+4c+3, so 64a+16b+4c=152, eqn D: 16a+4b+c=38.
B-A: E: 3a+b=7; D-A: 15a+3b=27; F: 5a+b=9
F-E: 2a=2, a=1, so b=4 and from A, c=11-5=6.
So f(x)=x^3+4x^2+6x+3=(x+1)(x^2+3x+3). This formula generates the series.
Let X=x+1, then f(X)=X((X-1)^2+3X) is an alternative equivalent formula.