# nth term

• May 19th 2013, 06:33 PM
roshanhero
nth term
How to find nth term of the sequence 6+13+24+39+.....
• May 19th 2013, 11:56 PM
chiro
Re: nth term
Hey roshanhero.

For your problem without any more clarification there will be an infinite number of sequences that fit your description.

You could find one by fitting a polynomial through them or use a program to generate a sequence.

Without any more constraints, a unique sequence is not really possible.
• May 20th 2013, 02:34 AM
Shakarri
Re: nth term
Your numbers seem random, there has to be a pattern if you want an expression for the nth term.
• May 20th 2013, 03:30 AM
ibdutt
Re: nth term
Notice that the difference between the successive terms is 7,11,15 etc and that is an AP
the sequence can be written as
6 + ( 6 + 7) + ( 6+7+11) + ( 6+7+11+15 ) and so on now i am sure you can get the nth term
• May 22nd 2013, 12:00 PM
HallsofIvy
Re: nth term
I would do this using "Newton's divided difference formula". The numbers are $a_0= 6$, $a_1= 13$, $a_2= 24$, and $a_3= 39$. The "First Differences" are $\Delta a_0= 13- 6= 7$, $\Delta a_1= 24- 13= 11$, $\Delta a_2= 39- 24= 15$ and then the "Second Differences" are $\Delta^2 a_0= 11- 7= 4$, $\Delta^2 a_1= 15- 11= 4$. If we assume that the sequence continues so that the second differences are alway four (and the third and higher differences are 0) we can write, in analogy with the Taylor's series, $a_n= a_0+ \Delta a_0 n+ (\Delta^2 a_0/2)n(n-1)= 6+ 7n+ (4/2)n(n- 1)= 6+ 7n+ 2n^2- 2n= 6+ 5n+ 2n^2$.

You will note that when n= 0, $a_0= 6+ 5(0)+ 2(0^2)= 6$, when n= 1, $a_1= 6+ 5(1)+ 2(1^2)= 13$, when n= 2, $a_2= 6+ 5(2)+ 2(2^2)= 24$, and when n= 3, $a_3= 6+ 5(3)+ 2(3^2)= 39$ exactly as we want.

You could also use the fact that, since the second differences are constant, the sequence is a quadratic function of n: $a_n= an^2+ bn+ c$. Then use $a_0= a(0)+ b(0)+ c= c= 6$,
$a_1= a(1^2)+ b(1)+ c= a+ b+ c= 13$ and $a_2= a(2^2)+ b(2)c= 4a+ 2b+ c= 24$ to solve for a, b, and c.