Question

Let f(k) be the sum of the cubes of all natural numbers less than or equal to k. How many natural numbers k exist such that f(k) is a perfect power of 3?

A

0

B

1

C

2

D

3

Hi,

We know that f(k) is the sum of the cubes of the numbers.

f(k)=[k(k+1)/2]^2 = 3^n

=> k^2 X (k+1)^2 = 3^n X 2^2

So, either k or k+1 has to be equal to 2. If k+1 = 2, k=1. in this case f(k) won't be a perfect power of 3.

If k=2, k+1 =3 and f(k) will be a perfect power of 3. 

Thus, the answer to this question is B.