ok so I have the numbers 1,2,6,22,86....or 1, 2^(1/2), 6^(1/2), 22^(1/2)....
Now heres what I know about them.. there is a diffrence of 4^n between them..
Also I noticed a pattern when I start with and S1=2 and make the formula 2+4n. Then I wanted to say that if a sequence Sn= 2+4n. Then the subsequence of S(Sn) if you put S(s1=2) I get 6 and S(s2=6) I get 22... however I need a cleaner formula I'm just stuck on this one piece then I can do the proof.
Now heres what I know about them.. there is a diffrence of 4^n between them..
Also I noticed a pattern when I start with and S1=2 and make the formula 2+4n. Then I wanted to say that if a sequence Sn= 2+4n. Then the subsequence of S(Sn) if you put S(s1=2) I get 6 and S(s2=6) I get 22... however I need a cleaner formula I'm just stuck on this one piece then I can do the proof.
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This looks like the successive differences are 4^n, as you said. In that case:
a_n+1 = a_n + 4^(n-1)
You can put this in closed form as:
a_n = (1/3) 4^(n-1) + 2/3
Ed: You're welcome... when you take Calculus, you'll learn some tricks that help with these problems. For example, I knew that the answer had to be in the form B*4^n + C, and I just had to solve for B and C:
B 4^(n) + C = B 4^(n-1) + C + 4^(n-1)
B 4^(n) = (B+1) 4^(n-1)
4B = B + 1
B = 1/3 - then just solve for C.
a_n+1 = a_n + 4^(n-1)
You can put this in closed form as:
a_n = (1/3) 4^(n-1) + 2/3
Ed: You're welcome... when you take Calculus, you'll learn some tricks that help with these problems. For example, I knew that the answer had to be in the form B*4^n + C, and I just had to solve for B and C:
B 4^(n) + C = B 4^(n-1) + C + 4^(n-1)
B 4^(n) = (B+1) 4^(n-1)
4B = B + 1
B = 1/3 - then just solve for C.