Another Math Problem...

[digitalmaster287]

Can someone help me find the formula for the nth term in the following sequence?

1, 2, 5, 14, 41...

I know the differences triple each time (add 1 then add 3 then 9 etc.) but I cant find the formula for the nth term. Is it possible to do?

 

[Doc_Holiday]

Multiply by 3 and add to the number before it.

 

[Kada]

Isn't it just multiply by 3, and then subract 1?

 

[digitalmaster287]

Thats just to find the next number in the sequence tho. Im looking for the forumla that lets you get the value of any number in the sequence based on its place.

Like:

2, 5, 8, 11

In that sequence, the formula is 3n- 1so if I wanted to find the 100th number, it would be 3(100) -1 or 299

 

[Mcscruff]

Is it a geometric sequence though?

if it is you can use:

An=(a1)(r)^(n-1)

basically nth term is equal to 1st term multiplied by common ratio raised to power of n-1


hope that helps..

 

[digitalmaster287]

Thats what I was thinkingbut its not really ageometric sequence, because youre not just multiplying the previous number in the chain by 3, youre multiplying it by three AND subtracting one. Im not sure how to alter the formula to do that though.

 

[moogle]

Yay, usually I get to the Proom math topics too late.

It's a summation of a geometric sequence, with a little offset. The formula for that looks something like:

3^n - 1
--------- = a_n
3 - 1

a_0 = (3^0 - 1)/2 = 0
a_1 = (3^1 - 1)/2 = 1
a_2 = (3^2 - 1)/2 = 4
a_3 = (3^3 - 1)/2 = 13
a_4 = (3^4 - 1)/2 = 40

then uhh.. just add 1 to get your sequence.

so you get

(3^n - 1) / 2 + 1 = (3^n + 1) / 2

or if you wanna index it so a_1 is the first term, and not a_0, then...

[3^(n-1) + 1] / 2 would be the answer.

 

[digitalmaster287]

Thanks man. That was bugging me for a while.

 

[Bahamut]

Props to moogle. It's a good thing HE helped you, because I suck at this kind of stuff.

 

[Professor Pro]

i think theres websites you can go on google and find the awnser to those sort of questions.