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: Write $T_n$ using the sum of cubes formula.
$T_n = 1^3 + 2^3 + \cdots + n^3 = \left(\dfrac{n(n+1)}{2}\right)^2$, so every $T_n$ is a perfect square. Powers of 2 that are also perfect squares are $1, 4, 16, 64, 256, \ldots$, that is, $2^0, 2^2, 2^4, 2^6, \ldots$ (even exponents only).

Step 2: Compute the first several values of $T_n$ directly.
$T_1 = 1$
$T_2 = 1+8 = 9$
$T_3 = 9+27 = 36$
$T_4 = 36+64 = 100$
$T_5 = 100+125 = 225$
$T_6 = 225+216 = 441$
None of $9, 36, 100, 225, 441$ is a power of 2, since they equal $3^2, 6^2, 10^2, 15^2, 21^2$, and 3, 6, 10, 15, 21 are not powers of 2. Only $T_1=1=2^0$ shows up as a power of 2, and its exponent is 0.

Step 3: Explain why this pattern never changes for larger n.
$T_n$ being a power of 2 needs $t_n = n(n+1)/2$ to be a power of 2, since $T_n = t_n^2$ and squares of powers of 2 are powers of 2, and conversely. n and n+1 never share a common factor, so if their product $n(n+1) = 2 t_n$ is to have only 2 as a prime factor, both n and n+1 must separately be powers of 2.
Powers of 2 spread out fast: after 1 and 2, the gaps are 2, 4, 8, 16 and so on, so no two powers of 2 beyond $\{1,2\}$ are ever consecutive integers. This confirms $n=1$ is the only place $t_n$ can be a power of 2, and it only gives $2^0$.

Step 4: Apply the positive power condition.
The question specifically asks for a positive power of 2, meaning $2^k$ with $k \geq 1$, values like 2, 4, 8, 16 and so on. Since the one case found, $T_1 = 1 = 2^0$, has exponent 0, it does not count.
No other n gets close, as shown by both the direct computation for small n and the coprimality argument for all n.

Step 5: State the conclusion.
So there is no n at all for which $T_n$ works out to a positive power of 2.
\[ \boxed{\text{No such n exists, so statement (A) is correct}} \]