Question

For \(n = 1, 2, 3, \ldots\), let \(T_n = 1^3 + 2^3 + \cdots + n^3\). Which one of the following statements is correct?

• A. There is no value of \(n\) for which \(T_n\) is a positive power of 2.
• B. There is exactly one value of \(n\) for which \(T_n\) is a positive power of 2.
• C. There are exactly two values of \(n\) for which \(T_n\) is a positive power of 2.
• D. There are more than two values of \(n\) for which \(T_n\) is a positive power of 2.

Hint:

Use T_n = (n(n+1)/2)^2 and note that n and n+1 are coprime, so both would have to be powers of 2 themselves for T_n to be a power of 2.

Answer 1

The Correct Option is A

Step 1: Recall the formula for the sum of cubes.
The sum of the first n cubes has a well known closed form:
\[ T_n = 1^3+2^3+\cdots+n^3 = \left(\dfrac{n(n+1)}{2}\right)^2 \]
So \(T_n\) is always a perfect square, the square of the nth triangular number \(t_n = \dfrac{n(n+1)}{2}\).

Step 2: Turn the condition into one about \(t_n\).
A positive power of 2 means \(T_n = 2^k\) for some integer \(k \geq 1\). Since \(T_n = t_n^2\), we need:
\[ t_n^2 = 2^k \]
For a perfect square to equal a power of 2, the exponent k must be even, and \(t_n\) itself must be a power of 2. Write \(k = 2m\), so we need:
\[ t_n = \dfrac{n(n+1)}{2} = 2^m \]
for some integer \(m \geq 0\) (and we need \(m \geq 1\) at the end, to make \(k=2m\) positive).

Step 3: Use that n and n+1 share no common factor.
n and n+1 are consecutive integers, so \(\gcd(n, n+1) = 1\), they share no factors at all.
From \(n(n+1) = 2 \cdot 2^m = 2^{m+1}\), the whole product is a power of 2. Since n and n+1 have no common factor, and their product only has 2 as a prime factor, each of n and n+1 must individually be a power of 2 (one of them could be \(2^0=1\)).

Step 4: Find which consecutive integers are both powers of 2.
Powers of 2 are 1, 2, 4, 8, 16, 32, and so on. Consecutive powers of 2 get further apart as they grow (4 and 8 differ by 4, 8 and 16 differ by 8), so the only pair of powers of 2 that are consecutive integers is 1 and 2.
This means the only candidate is \(n = 1\) and \(n+1 = 2\).

Step 5: Check what this gives for \(T_n\).
At \(n=1\), \(t_1 = \dfrac{1 \cdot 2}{2} = 1 = 2^0\), so \(m = 0\), giving \(k = 2m = 0\).
That means \(T_1 = 1^3 = 1 = 2^0\), which is a power of 2, but the exponent is 0, not a positive integer. The question asks for a positive power of 2, so \(k \geq 1\) is required, and \(T_1 = 2^0\) does not qualify.
Since \(n=1\) was the only candidate at all, and even that one fails the positive power requirement, there is no value of n that works.

Final Answer:
There is no value of n for which \(T_n\) is a positive power of 2. \[ \boxed{\text{Statement (A) is correct}} \]