# Nth term

1. May 19, 2013

### roshan2004

How to find nth term of the sequence 6+13+24+39+.....

2. May 19, 2013

### Simon Bridge

You can't. There is no unique sequence longer than 4 giving rise to that set of numbers.
You need to make some sort of assumption about what sort of sequence you have there.

i.e. if you put x0=6, x1=13, x2=24, and x3=39; then the nth term could be (2n)6: n=even, and 13n: n= odd. BUt if I'm going to do that, then I could just say that x(n) = x(n-2)+18: n=even, x(n)=x(n-2)+26: n=odd ;)

note: what have the + signs to do with it?

Last edited: May 19, 2013
3. May 19, 2013

### roshan2004

It is actually 6,13,24,39......Their differences are in Arithmetic progression

4. May 19, 2013

### Simon Bridge

It is only "actually" if you restrict possible generation rules to a limited subset.
Or do you mean that you want to know how to write that particular rule out in some compact way?
Please make the question clear - you appear to be relying on a context we don't have.

Perhaps you want the sequence generation rule in terms of something like x(n)=... ?

5. May 20, 2013

### HallsofIvy

Staff Emeritus
Any reason why you didn't tell us that before? As Simon Bridge said, there are infinitely many different sequences that start out "6, 13, 24, 39". We might have noticed that 13- 6= 7, 24- 13= 11, 39- 24= 15 so that the "second differences" are 11- 7= 4 and 15- 11= 4 but there is no way of knowing that is always true without being told.

Knowing that the first differences are always 4 (so second and succeeding differences are 0) we can use "Newton's divided difference formula" which is very similar to "Taylor's series" for continuous functions. If a sequence, ${a_n}$ has first differences $\Delta_n$, second differences $\Delta^2_n$, etc. then $a_n= a_0+ \Delta_0 n+ (\Delta_0^2/2) n(n-1)+ \cdot\cdot\cdot+ (\Delta_0^i/i!)n(n-1)(n-2)\cdot\cdot\cdot(n- i- 1)$

In this case, with $a_0= 6$, $a_1= 13$, $a_2= 24$, $a_3= 39$, $\Delta_0= 7$, $\Delta_1= 11$, $\Delta_2= 15$, and continuing so that $\Delta^2_i= 4$ for all i, then all higher differences are 0 and Newton's formula becomes
$$6+ 7n+ (4/2)n(n-1)= 6+ 7n+ 2n^2- 2n= 2n^2+ 5n+ 6$$

That is $a_0= 2(0^2)+ 5(0)+ 6= 6$
$a_1= 2(1^2)+ 5(1)+ 6= 2+ 5+ 6= 13$,
$a_2= 2(2^2)+ 5(2)+ 6= 8+ 10+ 6= 24$,
and $a_3= 2(3^2)+ 5(3)+ 6= 19+ 15+ 6= 39$ as desired.

Notice that, because I chose to simplify by starting the sum at i= 0, the "nth term" is
$2(n-1)^2+ 5(n-1)+ 6= 2n^2- 4n+ 2+ 5n- 5+ 6= 2n^2+ n+ 3$
so that $a_1= 2(1)+ 1+ 3= 6$, $a_2= 2(4)+ 2+ 3= 13$, $a_3= 2(9)+ 3+ 3= 24$, and $a_4= 2(16)+ 4+ 3= 39.$

You could also have done this by recognizing from the start that (as a result of Newton's divided difference formula) since the second differences are constant, the formula must be quadratic in n:
$an^2+ bn+ c$ and then use the values given to find a, b, and c.

Last edited: May 20, 2013
6. May 20, 2013

### Stephen Tashi

Do you mean "nth term of the sequence" 6,13,24,39,... or "nth partial sum of the series" 6+13+24+39...?