SOLUTION: Help with special sequences? I'm stuck on these two questions and it would be greatly appreciated if you could help me! Even if its one question, I would really like to get the wh

Algebra ->  Sequences-and-series -> SOLUTION: Help with special sequences? I'm stuck on these two questions and it would be greatly appreciated if you could help me! Even if its one question, I would really like to get the wh      Log On


   



Question 1005810: Help with special sequences?
I'm stuck on these two questions and it would be greatly appreciated if you could help me! Even if its one question, I would really like to get the whole concept to help prepare for my test! Thank you!!
1. Find the 4th term if the sequence in which a^1 = 2 and a^n+1 = -4a^n + 2.
2.Find the third iterate, x^3, of the function f(x) = 3x + 5 for an initial value of x^0 = 1.

Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
If a%5Bn%2B1%5D=-4a%5Bn%2B2%5D , the common ratio of consecutive terms is
r=a%5Bn%2B2%5D%2Fa%5Bn%2B1%5D=a%5Bn%2B2%5D%2F%28-4a%5Bn%2B2%5D%29=-1%2F4 .
It is a geometric sequence, where each term is the one before multiplied by r=-1%2F4 ,
so as for all geometric sequences, a%5Bn%5D=a%5B1%5D%2Ar%5E%28n-1%29.
In particular,
a%5B4%5D=a%5B1%5D%2Ar%5E%284-1%29=a%5B1%5D%2Ar%5E3 .
Since a%5B1%5D=2 and r=-1%2F4 ,
a%5B4%5D=2%2A%28-1%2F4%29%5E3=2%2A%28-1%2F64%29=highlight%28-1%2F32%29 .

The x%5Bn%5D terms are iterations using f%28x%29=3x%2B5
x%5B0%5D=1
x%5B1%5D=3%2Ax%5B0%5D%2B5=3%2A1%2B5=3%2B5=8
x%5B2%5D=3%2Ax%5B1%5D%2B5=3%2A8%2B5=24%2B5=29
x%5B3%5D=3%2Ax%5B2%5D%2B5=3%2A29%2B5=87%2B5=highlight%2892%29
Because the problem asked for the term x%5B3%5D , with index 3,
it was easy enough to calculate each of the terms from the previous term until I got to x%5B3%5D .
No need to get more complicated.
If the problem had asked for a term with a much higher index, such as the 10th term,
I would have found a "formula" to get that term directly, without calculating all the terms in between.
x%5B1%5D=3%2Ax%5B0%5D%2B5
x%5B2%5D=3%283%2Ax%5B0%5D%2B5%29%2B5=x%5B0%5D%2A3%5E2%2B5%2A3%2B5
x%5B3%5D=3%28x%5B0%5D%2A3%5E2%2B5%2A3%2B5%29%2B5=x%5B0%5D%2A3%5E3%2B5%2A3%5E2%2B5%2A3%2B5
%22.......................%22
x%5Bn%5D=x%5B0%5D%2A3%5En%2B5%2A3%5E%28n-1%29%2B5%2A3%5E%28n-2%29%2B%22...%22%2B5%2A3%2B5
x%5Bn%5D=x%5B0%5D%2A3%5En%2B5%283%5E%28n-1%29%2B3%5E%28n-2%29%2B%22...%22%2B3%2B1%29
Since 3%5E%28n-1%29%2B3%5E%28n-2%29%2B%22...%22%2B3%2B1 is the sum of a geometric sequence with first term 1 and common ratio r=3 ,
that sum is 3%5E%28n-1%29%2B3%5E%28n-2%29%2B%22...%22%2B3%2B1=%283%5En-1%29%283-1%29 , so
x%5Bn%5D=x%5B0%5D%2A3%5En%2B5%28%283%5En-1%29%2F%283-1%29%29
x%5Bn%5D=x%5B0%5D%2A3%5En%2B5%283%5En-1%29%2F2%29
That last "formula" applied to n=3 , and with x%5B0%5D=1 , yields
.