SOLUTION: a1 = 24^1/3, an+1=(an+24)^1/3, n ≥ 1. Then the integer part of a100 equals A. 2 B. 10 C. 100 D. 24

Algebra ->  Finance -> SOLUTION: a1 = 24^1/3, an+1=(an+24)^1/3, n ≥ 1. Then the integer part of a100 equals A. 2 B. 10 C. 100 D. 24       Log On


   

Question 872658: a1 = 24^1/3, an+1=(an+24)^1/3, n ≥ 1. Then the integer part of a100 equals
A. 2
B. 10
C. 100
D. 24
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
8%3C24%3C27 <---> 2=8%5E%28%221+%2F+3%22%29%3C24%5E%28%221+%2F+3%22%29%3C27%5E%28%221+%2F+3%22%29=3
So 2%3Ca%5B1%5D%3C3 meaning that 2%3Ca%5Bn%5D%3C3 is true for n=1
Could it be true for all positive integer n values.
If it were true for n=h , we would know that 0%3Ca%5Bh%5D%3C3 .
Then 24%3Ca%5Bh%5D%2B24%3C3%2B24
24%3C%28a%5Bh%5D%2B24%29%3C27
24%5E%281%2F3%29%3C%28a%5Bh%5D%2B24%29%5E%281%2F3%29%3C27%5E%281%2F3%29=3 ,
and since 24%5E%28%221+%2F+3%22%29=a%5B1%5D and %28a%5Bh%5D%2B24%29%5E%281%2F3%29=a%5Bh%2B1%5D ,
we conclude that 2%3Ca%5B1%5D%3Ca%5Bh%2B1%5D%3C3 .
Since 2%3Ca%5Bn%5D%3C3 is true for n=1 ,
and being true for n=h makes it true for n=h%2B1 ,
then 2%3Ca%5Bn%5D%3C3 is true for all values of n .
The integer part of a%5Bn%5D for any n is 2 ,
and so the integer part of a%5B100%5D is highlight%282%29 .
It is true that a%5Bn%5D gets ever closer to 3 , but it can never get there.