A certain sequence starts with term_1

Question

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

* Kudos for all correct solutions

0akshay0 · Accepted Answer

term_n = 16^(2n - 1) = 2^{8n-4} term_1 = 2^{8-4} = 2^4 term_2 = 2^{16-4} = 2^{12} term_3 = 2^{24-4} = 2^{20} . . . term_k = 2^{8k-4} product of the first k terms of the sequence is 2^1600 term_1 * term_2 * term_3 ......* term_k = 2^{1600} 2^4*2^{12}*2^{20}.....*2^{8k-4}=2^{1600} 2^{4+12+20+....+8k-4}=2^{1600} 4+12+20+...8k-4 = 1600 sum of the above AP = (k/2)(4+8k-4) = 4K^2 4K^2=1600 K^2=400 k=20 Hence option C is correct Hit Kudos if you liked it 8-)

BrentGMATPrepNow · Answer

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

quantumliner · Answer

term_1 = 16 ^ (2-1) = 16 ^ 1
term_2 = 16 ^ (4-1) = 16 ^ 3
term_3 = 16 ^ (6-1) = 16 ^ 5

So we see a pattern forming.

question says that (16 ^ 1) * (16 ^ 3) * (16 ^ 5) *....   = 2^1600 = 16 ^ 400

hence the powers of 16 on the left side add up to 400

1+3+5 +....2k-1 = 400

Using the formula :: n/2   = 400

n/2  = 400 ==> n/2  = 400 ==> n^2 = 20^2 ==> n=20

Answer is C) 20

achuepnlx · Answer

Since,

nth term = term_n = 16^(2n - 1), the series will be 16,16^3,16^5,16^7,.....

Product of 2 terms = 16^4
Product of 3 terms = 16^9
Product of 4 terms = 16^16
Product of n terms = 16^n^2

2^1600 = 16^400 
ie; n^2 = 400, so n or here k = 20

Cez005 · Answer

Thanks for the post. Can you go over this part?

sum of the above AP = (k/2)(4+8k-4) =  4K^2

matt882 · Answer

-------------------------------------------------------------------------
Formula: #terms(Val_max+Val_min)/2 = sum of equidistant numbers

jhueglin · Answer

How do I get the formula : n/2   = 400 ???
Is there any previous post that explains this formula?

Pritishd · Answer

Hi  jhueglin

It is formula for the sum of the first n terms of a arithmetic progression. We use this formula to find the sum of the sequence 1 + 3 + 5 + 9 + ... + n = 400, which is an arithmetic progression with difference of 2 between any two consecutive terms.

Hope this helps!

bumpbot · Answer

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A certain sequence starts with term_1

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