Understanding the Binomial Expansion and Constant Term

The problem asks us to find the number of positive integers $k$ such that the constant term in the binomial expansion of $(2x^3 + \frac{3}{x^k})^{12}$ is of the form $2^8 \cdot l$, where $l$ is an odd integer.

Key Concept: General Term in Binomial Expansion For a binomial expansion of the form $(a + b)^n$, the general term (or the $(r+1)^{th}$ term) is given by: $T_{r+1} = {^n C_r} a^{n-r} b^r$ In our case, $a = 2x^3$, $b = \frac{3}{x^k} = 3x^{-k}$, and $n = 12$.

Step-by-Step Derivation of the General Term

1. Apply the general term formula: Substituting the values of $a$, $b$, and $n$ into the formula: $T_{r+1} = {^{12} C_r} (2x^3)^{12-r} (3x^{-k})^r$ Explanation: We apply the standard formula to get the general structure of any term in the expansion.

2. Separate coefficients and powers of x: Now, we simplify the expression by separating the numerical coefficients and the terms involving $x$: $T_{r+1} = {^{12} C_r} \cdot 2^{12-r} \cdot (x^3)^{12-r} \cdot 3^r \cdot (x^{-k})^r$ Using the exponent rule $(a^m)^n = a^{mn}$: $T_{r+1} = {^{12} C_r} \cdot 2^{12-r} \cdot x^{3(12-r)} \cdot 3^r \cdot x^{-kr}$ $T_{r+1} = {^{12} C_r} \cdot 2^{12-r} \cdot 3^r \cdot x^{36 - 3r - kr}$ Explanation: We simplify the expression to isolate the numerical part and combine all powers of $x$. This makes it easier to identify the exponent of $x$.

Condition for a Constant Term

For a term to be a constant term, it must not contain $x$. This means the exponent of $x$ in the general term must be equal to zero.

1. Set the exponent of x to zero: $36 - 3r - kr = 0$ Explanation: This is the crucial condition for finding a constant term. If the exponent of $x$ is not zero, the term will involve $x$.

2. Rearrange to find the relationship between r and k: $36 = 3r + kr$ Factor out $r$: $36 = r(3 + k)$ Therefore, $k+3 = \frac{36}{r} \implies k = \frac{36}{r} - 3$ Explanation: We isolate $k$ to easily find its value for different possible values of $r$.

Constraints on r and k

• In a binomial expansion $(a+b)^n$, the value of $r$ can range from $0$ to $n$. Here, $n=12$, so $r \in \{0, 1, 2, \dots, 12\}$.
• The problem states that $k$ must be a positive integer. This means $k > 0$.
• Since $k = \frac{36}{r} - 3$, for $k$ to be an integer, $r$ must be a divisor of 36.

Finding Valid Integer Values of k

Let's test the values of $r$ (divisors of 36 between 1 and 12, inclusive, as $r=0$ would make the expression undefined) and determine the corresponding $k$:

• If $r=1$, $k = \frac{36}{1} - 3 = 33$
• If $r=2$, $k = \frac{36}{2} - 3 = 18 - 3 = 15$
• If $r=3$, $k = \frac{36}{3} - 3 = 12 - 3 = 9$
• If $r=4$, $k = \frac{36}{4} - 3 = 9 - 3 = 6$
• If $r=6$, $k = \frac{36}{6} - 3 = 6 - 3 = 3$
• If $r=9$, $k = \frac{36}{9} - 3 = 4 - 3 = 1$
• If $r=12$, $k = \frac{36}{12} - 3 = 3 - 3 = 0$

Since $k$ must be a positive integer, the possible values for $k$ are $33, 15, 9, 6, 3, 1$. The corresponding $(r, k)$ pairs are: $(1, 33), (2, 15), (3, 9), (4, 6), (6, 3), (9, 1)$.

Explanation: We systematically evaluate $k$ for each possible $r$ that yields an integer $k$. We then filter these to only include positive integer values for $k$.

Analyzing the Constant Term for each (r, k) pair

Now we need to find the constant term for each valid $(r, k)$ pair and check if it can be written as $2^8 \cdot l$, where $l$ is an odd integer. The constant term is given by the numerical part of $T_{r+1}$ when the exponent of $x$ is zero: $C = {^{12} C_r} \cdot 2^{12-r} \cdot 3^r$

Case 1: (r=1, k=33) $C = {^{12} C_1} \cdot 2^{12-1} \cdot 3^1 = 12 \cdot 2^{11} \cdot 3 = (2^2 \cdot 3) \cdot 2^{11} \cdot 3 = 2^{13} \cdot 3^2$ Here, the power of 2 is 13, which is not 8. So, $k=33$ is not valid.

Case 2: (r=2, k=15) $C = {^{12} C_2} \cdot 2^{12-2} \cdot 3^2 = \frac{12 \cdot 11}{2} \cdot 2^{10} \cdot 3^2 = 66 \cdot 2^{10} \cdot 3^2 = (2 \cdot 3 \cdot 11) \cdot 2^{10} \cdot 3^2 = 2^{11} \cdot 3^3 \cdot 11$ Here, the power of 2 is 11, which is not 8. So, $k=15$ is not valid.

Case 3: (r=3, k=9) $C = {^{12} C_3} \cdot 2^{12-3} \cdot 3^3 = \frac{12 \cdot 11 \cdot 10}{3 \cdot 2 \cdot 1} \cdot 2^9 \cdot 3^3 = (2 \cdot 11 \cdot 10) \cdot 2^9 \cdot 3^3 = 220 \cdot 2^9 \cdot 3^3$ $C = (2^2 \cdot 5 \cdot 11) \cdot 2^9 \cdot 3^3 = 2^{11} \cdot 5 \cdot 11 \cdot 3^3$ Here, the power of 2 is 11, which is not 8. So, $k=9$ is not valid.

Case 4: (r=4, k=6) $C = {^{12} C_4} \cdot 2^{12-4} \cdot 3^4 = \frac{12 \cdot 11 \cdot 10 \cdot 9}{4 \cdot 3 \cdot 2 \cdot 1} \cdot 2^8 \cdot 3^4 = (11 \cdot 5 \cdot 9) \cdot 2^8 \cdot 3^4$ $C = (495) \cdot 2^8 \cdot 3^4 = 2^8 \cdot (495 \cdot 3^4) = 2^8 \cdot (495 \cdot 81)$ $C = 2^8 \cdot 40095$ Here, the power of 2 is 8, and $l = 40095$, which is an odd integer. So, $k=6$ is a valid value.

Case 5: (r=6, k=3) $C = {^{12} C_6} \cdot 2^{12-6} \cdot 3^6 = \frac{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7}{6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} \cdot 2^6 \cdot 3^6 = (11 \cdot 2 \cdot 3 \cdot 2 \cdot 7) \cdot 2^6 \cdot 3^6$ $C = (924) \cdot 2^6 \cdot 3^6 = (2^2 \cdot 3^2 \cdot 7 \cdot 11) \cdot 2^6 \cdot 3^6 = 2^8 \cdot 3^8 \cdot 7 \cdot 11$ $C = 2^8 \cdot (3^8 \cdot 7 \cdot 11) = 2^8 \cdot (6561 \cdot 7 \cdot 11) = 2^8 \cdot (45927 \cdot 11) = 2^8 \cdot 505197$ Here, the power of 2 is 8, and $l = 505197$, which is an odd integer. So, $k=3$ is a valid value.

Case 6: (r=9, k=1) $C = {^{12} C_9} \cdot 2^{12-9} \cdot 3^9 = {^{12} C_3} \cdot 2^3 \cdot 3^9 = \frac{12 \cdot 11 \cdot 10}{3 \cdot 2 \cdot 1} \cdot 2^3 \cdot 3^9 = 220 \cdot 2^3 \cdot 3^9$ $C = (2^2 \cdot 5 \cdot 11) \cdot 2^3 \cdot 3^9 = 2^5 \cdot 5 \cdot 11 \cdot 3^9$ Here, the power of 2 is 5, which is not 8. So, $k=1$ is not valid.

Explanation: For each potential $k$, we calculate the constant term and then carefully factorize it to determine the exact power of 2. We then verify if the remaining factor $l$ is odd. This meticulous check ensures that both conditions ($2^8$ and odd $l$) are met.

Summary and Conclusion

From our analysis, only for $k=3$ (when $r=6$) and $k=6$ (when $r=4$) does the constant term satisfy the condition of being of the form $2^8 \cdot l$, where $l$ is an odd integer.

Therefore, there are 2 positive integer values of $k$ that satisfy the given condition.

Key Takeaway: When dealing with conditions involving the prime factorization of a term (like $2^8 \cdot l$), it's crucial to perform a complete prime factorization of the combinatorial coefficients and powers in the term to accurately determine the exponents of each prime factor. Do not assume the remaining part after extracting a power of 2 is odd without verification.