Rewritten Solution: Identifying and Summing Rational Terms in Binomial Expansions

1. Key Concept: The Binomial Theorem and Conditions for Rational Terms

This problem asks us to find the sum of all rational terms in a binomial expansion. The foundational concept here is the Binomial Theorem, which provides a formula for expanding expressions of the form $(a+b)^n$.

The Binomial Theorem states that for any real numbers $a$ and $b$, and any non-negative integer $n$: $(a+b)^n = \sum_{r=0}^{n} {^nC_r} a^{n-r} b^r$ Each individual term in this expansion is called the general term, often denoted as $T_{r+1}$ (the $(r+1)^{th}$ term), given by: $T_{r+1} = {^nC_r} a^{n-r} b^r$ where ${^nC_r} = \frac{n!}{r!(n-r)!}$ is the binomial coefficient, which is always an integer.

For a term $T_{r+1}$ to be a rational number, we need to consider the nature of $a$ and $b$. If $a$ and $b$ are expressed with fractional exponents (e.g., $a = x^{1/p}$ and $b = y^{1/q}$), then the general term becomes: $T_{r+1} = {^nC_r} (x^{1/p})^{n-r} (y^{1/q})^r = {^nC_r} x^{\frac{n-r}{p}} y^{\frac{r}{q}}$ Since ${^nC_r}$ is always an integer (and thus rational), the rationality of $T_{r+1}$ depends entirely on the terms $x^{\frac{n-r}{p}}$ and $y^{\frac{r}{q}}$. For the entire term to be rational, the exponents of $x$ and $y$ must resolve to non-negative integers. That is, both $\frac{n-r}{p}$ and $\frac{r}{q}$ must be non-negative integers. This implies that $(n-r)$ must be a multiple of $p$, and $r$ must be a multiple of $q$.

2. Step-by-Step Solution

Let's apply this understanding to the given problem: The expansion is $(2^{1/3} + 3^{1/4})^{12}$.

Here, we can identify:

• $a = 2^{1/3}$
• $b = 3^{1/4}$
• $n = 12$

Step 1: Write Down the General Term ($T_{r+1}$) Using the binomial theorem, the general term for this expansion is: $T_{r+1} = {^nC_r} a^{n-r} b^r$ Substitute the values of $n$, $a$, and $b$: $T_{r+1} = {^{12}C_r} (2^{1/3})^{12-r} (3^{1/4})^r$ Explanation: This is the direct application of the general term formula. We've substituted the specific components of our given binomial expression into the standard formula. The variable $r$ will take integer values from $0$ to $n$ (inclusive), determining each specific term in the expansion.

Step 2: Simplify the Exponents in the General Term Now, we simplify the powers using the exponent rule $(x^m)^k = x^{mk}$: $T_{r+1} = {^{12}C_r} 2^{\frac{12-r}{3}} 3^{\frac{r}{4}}$ Explanation: Simplifying the exponents makes it easier to analyze the conditions for rationality in the next step. By expressing the powers as fractions, we can directly see the denominators that must divide the numerators for the result to be an integer.

Step 3: Determine Conditions for Rationality For $T_{r+1}$ to be a rational number, two conditions must be met regarding the exponents of the base numbers (2 and 3):

1. The exponent of 2, which is $\frac{12-r}{3}$, must be a non-negative integer. This means $(12-r)$ must be a multiple of 3. Since 12 is already a multiple of 3, for $(12-r)$ to be a multiple of 3, $r$ must also be a multiple of 3. Condition 1: $r$ must be a multiple of 3.

2. The exponent of 3, which is $\frac{r}{4}$, must be a non-negative integer. This means $r$ must be a multiple of 4. Condition 2: $r$ must be a multiple of 4.

Explanation: The binomial coefficient ${^{12}C_r}$ is always an integer, so it's inherently rational. The potential for irrationality comes from the terms $2^{\dots}$ and $3^{\dots}$. For the entire term to be rational, these parts must resolve to rational numbers. Since 2 and 3 are prime numbers, this can only happen if their exponents are integers. We set up divisibility rules based on the denominators of the fractional exponents. The exponents must also be non-negative because $r$ is non-negative and $12-r$ will also be non-negative for $r \le 12$.

Step 4: Find Possible Values of $r$ The value of $r$ in the binomial expansion ranges from $0$ to $n$. In this problem, $0 \le r \le 12$. We need $r$ to satisfy both Condition 1 ($r$ is a multiple of 3) and Condition 2 ($r$ is a multiple of 4). This means $r$ must be a common multiple of 3 and 4. The least common multiple (LCM) of 3 and 4 is 12. Therefore, $r$ must be a multiple of 12.

Now, we check which multiples of 12 fall within the valid range for $r$ ($0 \le r \le 12$):

• If $r=0$, it is a multiple of 12.
• If $r=12$, it is a multiple of 12.
• Any other multiple of 12 (e.g., 24) would be outside the range $0 \le r \le 12$.

So, the only possible values of $r$ for which the terms are rational are $r=0$ and $r=12$.

Explanation: We combine the two necessary conditions. If $r$ must be divisible by both 3 and 4, it must be divisible by their LCM. Then, we apply the fundamental constraint that $r$ can only be an integer from $0$ to $n$. This effectively filters down the infinite list of common multiples to a finite, manageable set of values.

Step 5: Calculate the Rational Terms Now we substitute these valid values of $r$ back into the simplified general term formula: $T_{r+1} = {^{12}C_r} 2^{\frac{12-r}{3}} 3^{\frac{r}{4}}$

For $r=0$: This gives us the first term ($T_1$). $T_{0+1} = T_1 = {^{12}C_0} 2^{\frac{12-0}{3}} 3^{\frac{0}{4}}$ Recall that ${^{12}C_0} = 1$, $2^{12/3} = 2^4$, and $3^0 = 1$. $T_1 = 1 \times 2^4 \times 1$ $T_1 = 1 \times 16 \times 1$ $T_1 = 16$

For $r=12$: This gives us the thirteenth term ($T_{13}$). $T_{12+1} = T_{13} = {^{12}C_{12}} 2^{\frac{12-12}{3}} 3^{\frac{12}{4}}$ Recall that ${^{12}C_{12}} = 1$, $2^{0/3} = 2^0 = 1$, and $3^{12/4} = 3^3$. $T_{13} = 1 \times 2^0 \times 3^3$ $T_{13} = 1 \times 1 \times 27$ $T_{13} = 27$

Explanation: We've found the specific values of $r$ that yield rational terms. Now we simply plug these values back into the general term expression and perform the arithmetic. It's important to correctly evaluate the binomial coefficients and the powers.

Step 6: Sum the Rational Terms The problem asks for the sum of all rational terms. We have identified two such terms: $T_1 = 16$ and $T_{13} = 27$. Sum of rational terms $= T_1 + T_{13} = 16 + 27 = 43$.

Explanation: This is the final step, where we aggregate all the rational terms we've calculated to arrive at the solution requested by the question.

3. Tips and Common Mistakes to Avoid

• Range of $r$ is crucial: Always remember that $r$ must be an integer from $0$ to $n$ (inclusive). This is a common oversight that can lead to incorrect $r$ values.
• LCM, not just common multiples: When $r$ must be a multiple of multiple numbers (e.g., $x$ and $y$), it must be a multiple of their Least Common Multiple, $\text{LCM}(x, y)$. Don't just pick any common multiple.
• Binomial Coefficient is always rational: Remember that ${^nC_r}$ will always be an integer, so it never contributes to irrationality. Focus only on the bases with fractional exponents.
• Zero Exponent Rule: Any non-zero base raised to the power of 0 equals 1 (e.g., $X^0 = 1$). This often appears in terms where $r=0$ or $r=n$.
• Careful with Arithmetic: Double-check calculations, especially with exponents and binomial coefficients.

4. Summary and Key Takeaway

To find the sum of rational terms in a binomial expansion of the form $(x^{1/p} + y^{1/q})^n$:

1. Formulate the General Term: Write $T_{r+1} = {^nC_r} x^{\frac{n-r}{p}} y^{\frac{r}{q}}$.
2. Establish Rationality Conditions: For the term to be rational, the exponents $\frac{n-r}{p}$ and $\frac{r}{q}$ must both be non-negative integers. This leads to divisibility conditions for $r$.
3. Identify Valid $r$ Values: Find the values of $r$ (within the range $0 \le r \le n$) that satisfy all divisibility conditions. This often involves finding the LCM of the denominators of the fractional exponents.
4. Calculate Terms: Substitute each valid $r$ value back into the general term formula and compute the rational terms.
5. Sum Terms: Add all the calculated rational terms to obtain the final answer.

The final answer is $\boxed{\text{43}}$.