In terms of n, 3, 7, 13, 27, 53, 107

[ziggie125]

Homework Statement

Find the explicit formula c1=1, c2=2, cn+1 = (cn + cn-1)/2 ; n>2

Homework Equations

The Attempt at a Solution

c1 = 1
c2 = 2
c3 = 3/2
c4 = 7/4
c5 = 13/8
c6 = 27/16
c7 = 53/32
c8 = 107/64

I know the bottom term is 2^n-2. I cannot find what the top is. If anyone sees it let me know thx.

Also if you know general form of 1, 1, -1, -1, 1, 1, -1, -1 ...

Thx a lot

 

[Refraction]

The numerator kind of looks like $$a_{n+2}=2a_{n}+a_{n+1}$$ to me, so it would be like this:
c4 = (2*2) + 3 = 7
c5 = (2*3) + 7 = 13
c6 = (2*7) + 13 = 27
etc, it only works when n > 2 which seems to fit into what you mentioned as well. Hope that helps a bit!

 

[lanedance]

try multplying out explicitly for the first few terms - you should hopefully be able to pick up a pattern in terms of sums of powers of 2
$$3 = 1+2$$
$$7 = 1 +2+2^2$$
and follow on from there...

 

[ziggie125]

I was told you cannot use the terms before it in the equation, ie. cn-1*2 + (-1)^n

 

[lanedance]

i'm not sure i know what you mean? try this though, write the c_n in term of a new variable a_n, for n>2
[tex]c_1 = 1[\tex]<br /> [tex]c_2 = 2[\tex]<br /> [tex]c_3 = \frac{3}{2} = \frac{a_3}{2}[\tex]<br /> [tex]c_4 = \frac{7}{2^2} = \frac{a_4}{2^2}[\tex]<br /> <br /> so you want to find the a_n, with<br /> $$c_n = \frac{a_n}{2^{n-2}}[\tex]<br /> <br /> how about seeing if you can find a recursion relation for the a_n, using the original... will be similar to what refraction posted, but the form should help lead to a general solution.$$[/tex][/tex][/tex][/tex]