d*k* 2 years=$600
k=300/d
Let dollars amount invested in 3 years be x
x*k*3 years=2400 x=800/k
Plugging i k in the last formula, i got 8d/3

if xy=1, what is 2^(x+y)^2 divided by 2^(x-y)^2?

If after simplification we are left with 2^ x^2 + y^2 / 2^ x^2 - y^2 ..

Corrected to C

given xy =1
2^(x+y)^2 / 2^(x-y)^2

2^(x^2 + y^2 + 2xy) / 2^(x^2 + y^2 - 2xy)

2^x^2 * 2^y^2 * 2^2xy / 2^x^2 * 2^y^2 * 2^-2xy

Cancel out and we are left with 2^2xy / 2^-2xy

xy =1

4 * 4 = 16 therefore C

rchadha7 wrote:given xy =1
2^(x+y)^2 / 2^(x-y)^2

2^(x^2 + y^2 + 2xy) / 2^(x^2 + y^2 - 2xy)

2^x^2 * 2^y^2 * 2^2xy / 2^x^2 * 2^y^2 * 2^-2xy

Cancel out and we are left with 2^2xy / 2^-2xy

xy =1

4 * 4 = 16 therefore C

How can you know the answer choice is C without being given the terms for choice A and Choice B? --- Thanks

Can someone better explain this task and the solution. Thanx.

For every integer k from 1 to 10, inclusive, the kth term of a certain sequence is given by (-1)^k+1(1/2^k). If T is the sum of the first 10 terms in the sequence, then T is..

1 greater than 2
2 between 1 and 2
3 between 1/2 and 1
4 between 1/4 and 1/2
5 less than 1/4

Bronson wrote:Can someone better explain this task and the solution. Thanx.

For every integer k from 1 to 10, inclusive, the kth term of a certain sequence is given by (-1)^k+1(1/2^k). If T is the sum of the first 10 terms in the sequence, then T is..

1 greater than 2
2 between 1 and 2
3 between 1/2 and 1
4 between 1/4 and 1/2
5 less than 1/4

This is a sequence question.

We have a 10 term sequence. Each term is defined by the equation given and to solve for each term, you substitute the number of the term in for k.

For example, the first term (k=1) would be (-1)^(1+1) * (1/2^1) = (-1)^2 * (1/2) = (1)(1/2) = 1/2.

We're asked to solve for the sum of the first 10 terms, i.e. for the values k = 1, 2, 3, ..., 10

Let's start with the first half of the product:

(-1)^(k+1).

This part of the puzzle will alternate between +1 and -1, depending on if (K+1) is even or odd.

So, for k=1, it will equal +1 (since (-1)^(2) = 1)
for k=2, it will = -1
for k=3, it will = 1
for k=4, it will = -1
for k=5, it will = 1
for k=6, it will = -1
for k=7, it will =1
for k=8, it will = -1
for k=9, it will = 1
for k=10, it will = -1

The second part is just (1/2)^k, so the values will be 1/2, 1/4, 1/8, 1/16, ... The further we go, the less significant the number will be.

If we combine the first and second parts, we'll get:

k=1: (1)(1/2)
k=2: (-1)(1/4)
k=3: (1)(1/8)
k=4: (-1)(1/16)
and so on.

So, we'll end up with:

1/2 - 1/4 + 1/8 - 1/16 + 1/32 - 1/64 + 1/128 - 1/256 + 1/512 - 1/1024

If we look at the sum in pairs, we have:

(1/2 - 1/4) + (1/8 - 1/16) + (1/32 - 1/64) + (1/128 - 1/256) + (1/512 - 1/1024)

or:

1/4 + 1/16 + 1/64 + 1/256 + 1/1024

At a quick glance, we can narrow the answer down to either:

3) between 1/2 and 1
4) between 1/4 and 1/2

We could actually convert to a common denominator (1024) and solve, or we could reason it out. If we recognize that our terms are decreasing by 75% each time, we know that terms 2-5 will sum to less than 1/4. So, we'll end up with 1/4 + (less than 1/4) = less than 1/2, so answer choice (4) is correct.

Thank you Stuart!
You are the best.
Your explanations make problems so simple.