Problem Solving

lheiannie07 wrote:A certain sequence starts with term_1

For any term in the sequence, term_n = 16^(2n - 1)

If the PRODUCT of the first k terms of the sequence is 2^1600, what is the value of k?

A) 5
B) 10
C) 20
D) 40
E) 80

First notice that 2n - 1, will be ODD for all integer values of n. For example:
If n = 1, then 2n - 1 = 2(1) - 1 = 1
If n = 2, then 2n - 1 = 2(2) - 1 = 3
If n = 3, then 2n - 1 = 2(3) - 1 = 5
If n = 4, then 2n - 1 = 2(4) - 1 = 7
.
.
.
etc.


Now notice what happens when we add consecutive ODD numbers (starting with 1)
The first 1 ODD number: 1 = 1 (and 1 = 1²)
The first 2 ODD numbers: 1 + 3 = 4 (and 4 = 2²)
The first 3 ODD numbers: 1 + 3 + 5 = 9 (and 9 = 3²)
The first 4 ODD numbers: 1 + 3 + 5 + 7 = 16 (and 16 = 4²)
The first 5 ODD numbers: 1 + 3 + 5 + 7 + 9 = 25 (and 25 = 5²)
.
.
.
In general, the sum of the first k ODD numbers = k²

Now onto the question!!!

term_n = 16^(2n - 1)
term_1 = 16^(2(1) - 1) = 16^1
term_2 = 16^(2(2) - 1) = 16^3
term_3 = 16^(3(3) - 1) = 16^5
term_4 = 16^(2(4) - 1) = 16^7
etc

So, the PRODUCT of the first k terms = (16^1)(16^3)(16^5)(16^7)(16^9). . . (16^??)
When we multiply powers with the same base, we ADD the exponents.
So, the PRODUCT of the first k terms = 16^(1 + 3 + 5 + 7 + . . . ??)

Notice that the exponent here is equal to the SUM of the first k ODD numbers.
Well, we already know that the sum of the first k ODD numbers = k²
So, the PRODUCT of the first k terms = 16^(k²)

We're told that the PRODUCT of the first k terms is 2^1600
So, we can write: 16^(k²) = 2^1600

We need the same base, so let's rewrite 16 as 2^4
We get: (2^4)^(k²) = 2^1600
Apply power of a power law: 2^(4k²) = 2^1600
This means that 4k² = 1600
Divide both sides by 4 to get: k² = 400
Solve: k = 20 or -20
Since -20 makes no sense, we know that k = 20

In other words, the PRODUCT of the first 20 terms of the sequence is 2^1600,

Answer: C

Cheers,
Brent