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## Main Question or Discussion Point

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

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

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Simon Bridge

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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?

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?

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

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Simon Bridge

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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)=... ?

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HallsofIvy

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

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, [itex]{a_n}[/itex] has first differences [itex]\Delta_n[/itex], second differences [itex]\Delta^2_n[/itex], etc. then [itex]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)[/itex]

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

[tex]6+ 7n+ (4/2)n(n-1)= 6+ 7n+ 2n^2- 2n= 2n^2+ 5n+ 6[/tex]

That is [itex]a_0= 2(0^2)+ 5(0)+ 6= 6[/itex]

[itex]a_1= 2(1^2)+ 5(1)+ 6= 2+ 5+ 6= 13[/itex],

[itex]a_2= 2(2^2)+ 5(2)+ 6= 8+ 10+ 6= 24[/itex],

and [itex]a_3= 2(3^2)+ 5(3)+ 6= 19+ 15+ 6= 39[/itex] as desired.

Notice that, because I chose to simplify by starting the sum at i= 0, the "nth term" is

[itex]2(n-1)^2+ 5(n-1)+ 6= 2n^2- 4n+ 2+ 5n- 5+ 6= 2n^2+ n+ 3[/itex]

so that [itex]a_1= 2(1)+ 1+ 3= 6[/itex], [itex]a_2= 2(4)+ 2+ 3= 13[/itex], [itex]a_3= 2(9)+ 3+ 3= 24[/itex], and [itex]a_4= 2(16)+ 4+ 3= 39.[/itex]

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:

[itex]an^2+ bn+ c[/itex] and then use the values given to find a, b, and c.

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- #6

Stephen Tashi

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

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