Finding Number of Terms in {ak} Sequence - Help Needed

[murshid_islam]

this is the problem:
let {a_k} be a sequence of real numbers. and the sequence is defined as

$$a_0 = 2006$$

$$a_{n} = \log_{2}a_{n-1}$$

now i have to find out the number of terms in the sequence.

this is what i have done:
i see that a₅ is negative. so we can't find out a₆ or any other subsequent terms. so the number of terms is 6 (a₀ upto a₅).

but how can i find out the number of terms without calculating the terms upto a₅? is there a way to do it?

 

[Office_Shredder]

If $$a_{n} = \log_{2}a_{n-1}$$ is true, then $$2^{a_n} = a_{n-1)$$

So this tells us $$2^{2^{2...^{2^{a_n}}}} = a_0$$ So find how many times you can put two to itself before it goes past 2006.

I think that works

 

[tacman]

There is a way to find the number of terms in this sequence without calculating all the terms up to a5. This is known as finding the limit of a sequence, which is a concept in calculus. In this case, we can use the formula for the nth term of a geometric sequence, which is given by a_n = a_0 * r^(n-1), where a_0 is the first term and r is the common ratio.

In this sequence, we have a_0 = 2006 and a_n = log2(a_n-1). So, we can rewrite the formula as a_n = 2006 * (1/2)^(n-1). Now, to find the number of terms, we need to find the value of n that makes a_n = 0. Since we know that log2(a_n-1) cannot be negative, we can set a_n = 0 and solve for n.

0 = 2006 * (1/2)^(n-1)
0 = (1/2)^(n-1)
0 = 1/2

This means that n-1 = 0, and therefore, n = 1. So, the number of terms in this sequence is 1.

In general, to find the number of terms in a sequence, we need to find the value of n that makes the nth term equal to 0. This can be done by setting the nth term equal to 0 and solving for n using algebraic techniques.