# Thread: I need some help with sequence and series

1. ## I need some help with sequence and series

Well I'm stuck on this question from past 2 hours.Idk what to do with it.I have this sequence and I have to find a formula for it.

1,3,7,13,21,31.....

I know that the difference between every 2 terms is 2,4,6,8

so the difference between the terms can be just another sequence like this

summation of (2i) where i = 1 to n.

I need some help with it.

Thanks

2. Originally Posted by NidhiS
Well I'm stuck on this question from past 2 hours.Idk what to do with it.I have this sequence and I have to find a formula for it.

1,3,7,13,21,31.....

I know that the difference between every 2 terms is 2,4,6,8

so the difference between the terms can be just another sequence like this

summation of (2i) where i = 1 to n.

I need some help with it.

Thanks
note that these can be modeled with a quadratic function

how do we know that?

just look for when the difference between the terms become constant

$\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline & 1 & & 3 & & 7 & & 13 & & 21 & & 31 & \text{constant} \\
\hline \text{ first difference} & & 2 & & 4 & & 6 & & 8 & & 10 & & \text{linear} \\
\hline \text{ second difference}& & & 2 & & 2 & & 2 & & 2 & & & \text{quadratic} \\ \hline \end{array}$

stop since we get a constant difference of 2 between successive terms

so let $f(n) = an^2 + bn + c$ (the function that gives the nth term, starting at n = 0)

with
$f(0) = 1$ .............(1)
$f(1) = 3$ .............(2)
$f(2) = 7$ .............(3)

thus you have a system with 3 equations and 3 unknowns, you can solve for $a,~b, \mbox{ and }c$

3. er this doesn't work.I solved for a,b and c and i get
a=1
b=1
c=1
and the expression evaluates to n^2+n+1
so summation of this expression from n=1 to m

doesn't give the same series

Thanks for trying tho

4. sorry this really works.thanks!
you're cool.

5. Originally Posted by NidhiS
er this doesn't work.I solved for a,b and c and i get
a=1
b=1
c=1
and the expression evaluates to n^2+n+1
so summation of this expression from n=1 to m

doesn't give the same series

Thanks for trying tho
i never said summing n from 1 to m of this formula gives the sequence. i said this is a formula that gives the terms of the sequence. isn't that what you are after? a formula to get 1,3,7,13,21,31, ... ?

with this function, f(0) = 1, f(1) = 3, f(2) = 7, f(3) = 13, f(4) = 21, f(5) = 31, f(6) = you guessed it, 43, and so on

6. er yea,sorry.It does work thanks!

I didn't know how this sequence thing works and my book doesn't cover this stuff at all but it's still on the hw.Do you know how shd I go abt it.Lets just say I need soem startup ideas.

7. Originally Posted by NidhiS
er yea,sorry.It does work thanks!

I didn't know how this sequence thing works and my book doesn't cover this stuff at all but it's still on the hw.Do you know how shd I go abt it.Lets just say I need soem startup ideas.
well, i do not know what your homework covers, but as far as questions of this type go, where you have a sequence where the terms follow some predictable pattern with a kind of linear order, then modeling with a polynomial is usually the way to go. i have to know more about what you are doing to give an general tips. the most i can tell you now is to use google and wikipedia to look these things up

8. for instance

I have this naother sequence here

2,9,28,65,126

I can't even see a pattern here .I tried everything

9. lets just say it's n^3+1

10. Originally Posted by NidhiS
for instance

I have this naother sequence here

2,9,28,65,126

I can't even see a pattern here .I tried everything
yeah, i don't really see a pattern at the moment either. so lets just hit this guy with a huge polynomial to get all the terms.

we want a polynomial such that f(0) = 2, f(1) = 9, f(2) = 28, f(3) = 65, and f(4) = 128

so we go up to 5 terms, which is the number of terms in a 4th degree polynomial

so let $f(n) = an^4 + bn^3 + cn^2 + dn + e$

and continue

11. Originally Posted by NidhiS
lets just say it's n^3+1
it isn't. 9^3 + 1 is not 28. 28^3 + 1 is definitely not 65 and 65^3 + 1 is ... oh, you get my drift

12. OMGEEE it's 3n+1

13. Originally Posted by NidhiS
OMGEEE it's 3n+1
again, no. 3(28) + 1 is not 65, etc

it will not be a linear function. remember, if it were, we would have gotten a constant difference after the first try

here's a tip. stop trying to figure out a pattern from the first 2 or 3 elements only! what you say must work for everyone represented here

14. well it's n^3+1

here:

(1)^3+1 = 2
(2)^3 +1 =9
(3)^3 + 1 = 28
(4)^3 + 1 = 65
(5)^3 + 1 =126

I was right the first time.

er I'm stuck on this now 1,1/5 ,1/13,1/29 , 1/61

15. Originally Posted by NidhiS
well it's n^3+1

here:

(1)^3+1 = 2
(2)^3 +1 =9
(3)^3 + 1 = 28
(4)^3 + 1 = 65
(5)^3 + 1 =126

I was right the first time.
oh, ok, yes. you were right. i was imagining something else. i was thinking recursively. thinking we got to the next term by doing something to the first. ok. good job

er I'm stuck on this now 1,1/5 ,1/13,1/29 , 1/61
worry about the denominators. the sequence for the denominators is 1, 5, 13, 29, ... if you can figure that out, you are done, because you can just put one over that formula to get the sequence you want. can you figure that one out?

Page 1 of 2 12 Last