# Thread: Swamped by a junior series (-2, 5, 24, 61, 122 ... ) formula?

1. ## Disregard

Can anyone give me a formula for the series: -2, 5, 24, 61, 122 ....
This is more of an association problem which shows my brains are fried. I'm a "has been" in the field and this posting is a total failure of my part. It is a pre Calculus 11th (junior level) grade. Got it Npower 3 -3

2. ## Re: Disregard

nth term is n^3-3

3. ## Re: Swamped by a junior series (-2, 5, 24, 61, 122 ... ) formula?

For those of us who aren't as sharp as biffboy, here is an "algorithmic" approach- write these as $a_0= -2$, $a_1= 5$, $a_2= 24$, $a_3= 61$, $a_4= 122$, the "first differences" are $\Delta_0= 5- (-2)= 7$, $\Delta_1= 24- 5= 19$, $\Delta_2= 61- 24= 37$, and $\Delta_3= 122- 61= 61$; the "second differences" are $\Delta^2_0= 19- 7= 12$, $\Delta^2_1= 37-19= 18$, and $\Delta^2_1= 61- 37= 24$; the "third differences" are $\Delta^3_0= 18- 12= 6$, $\Delta^3_1= 6$, and $\Delta^2_2= 6$. Since the third differences are all the same, all succeeding differences are 0 and we can write this as a cubic equation.

"Newton's divided difference formula" is similar to the Taylor's series but with finite differences rather than derivatives: $f(0)+ \Delta f(0) n+ (\Delta^2 f(0)/2!)n(n-1)+ (\Delta^3 f(0)/3!)n(n-1)(n-2)+ \cdot\cdot\cdot$.

Here, that gives $-2+ 7n+ (12/2)n(n-1)+ (6/6)n(n-1)(n-2)= -2+ 7n+ 6n^2- 6n+ n^3- 3n^2+ 2n$
$n^3+ 3n^2+ 3n- 2= (n+1)^3- 3$.

There I started the indexing with n=0. Had I, instead, started with m= 1, so that n+1= m, that would be $m^3- 3$.

Great help!