If $x is invested at a constant annually compound interest..

[BTGmoderatorLU]

Is $x is invested at a constant annually compound interest rate of k percent, what is the ratio of the total amount including interest after 4n years to that after 3n years?

$$A.\ \left(1+\frac{k}{100}\right)^n$$
$$B.\ \left(1+k\right)^n$$
$$C.\ \left(1+\frac{kn}{100}\right)$$
$$D.\ \left(1+\frac{n}{100}\right)^k$$
$$E.\ \left(1+kn\right)$$

The OA is A.

I'm really confused with this PS question. Experts, any suggestion? I don't know how can I solve it. Thanks in advance.

[GMATGuruNY]

Let x = $1, k = 100%, and n = 2.
Since the amount in the account increases by 100% each year, it DOUBLES each year, as follows:
1, 2, 4, 8, 10, 12, 24, 48.
As the list above illustrates:
After 3n=6 years, the amount in the account = 12.
After 4n=8 years, the amount in the account = 48.
Thus:
(8th year)/(6th year) = 48/12 = 4. This is our target.

Now plug k=100 and n=2 into the answers to see which yields our target of 4.
Only A works:
(1+k/100)^n = (1 + 100/100)² = (1+1)² = 4.

The correct answer is A.

[Scott@TargetTestPrep]

Is $x is invested at a constant annually compound interest rate of k percent, what is the ratio of the total amount including interest after 4n years to that after 3n years?

$$A.\ \left(1+\frac{k}{100}\right)^n$$
$$B.\ \left(1+k\right)^n$$
$$C.\ \left(1+\frac{kn}{100}\right)$$
$$D.\ \left(1+\frac{n}{100}\right)^k$$
$$E.\ \left(1+kn\right)$$

After 4n years, the future value is x(1 + k/100)^4n and after 3 years, the future value is x(1 + k/100)^3n, so the ratio is:

[x(1 + k/100)^4n]/[x(1 + k/100)^3n] = (1 + k/100)^n

Answer: A