The Sum of All Rational Terms in a Binomial Expansion

This problem requires us to find the sum of all terms that are rational numbers in the expansion of $(2+\sqrt{3})^8$. To achieve this, we will utilize the Binomial Theorem and analyze the nature of each term.

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1. Key Concept: The Binomial Theorem and General Term

The Binomial Theorem provides a formula for expanding any binomial raised to a non-negative integer power. For an expression of the form $(a+b)^n$, the expansion is given by: $(a+b)^n = \sum_{r=0}^{n} {}^n C_r a^{n-r} b^r$ Here, ${}^n C_r = \frac{n!}{r!(n-r)!}$ is the binomial coefficient, which represents the number of ways to choose $r$ items from a set of $n$ items. The index $r$ ranges from $0$ to $n$.

For our specific problem, we have $(2+\sqrt{3})^8$. Comparing this with $(a+b)^n$:

• $a = 2$
• $b = \sqrt{3}$
• $n = 8$

The general term in this expansion, denoted as $T_{r+1}$ (since $r$ starts from $0$), is given by substituting these values into the general formula: $T_{r+1} = {}^8 C_r (2)^{8-r} (\sqrt{3})^r$ To simplify the term involving the square root, we can rewrite $(\sqrt{3})^r$ as $(3^{1/2})^r = 3^{r/2}$. Therefore, the general term becomes: $T_{r+1} = {}^8 C_r (2)^{8-r} (3)^{r/2}$ Why this step is taken: Writing down the general term is crucial because it allows us to examine the structure of every term in the expansion systematically. By analyzing this single expression, we can determine the conditions under which a term will be rational or irrational, without having to expand the entire expression.

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2. Condition for Rational Terms

A term in the binomial expansion is considered rational if it can be expressed as a ratio of two integers (i.e., it does not contain any irrational components like $\sqrt{3}$). Let's analyze each factor in our general term $T_{r+1} = {}^8 C_r (2)^{8-r} (3)^{r/2}$:

• The binomial coefficient ${}^8 C_r$: By definition, ${}^8 C_r$ is always an integer for integer values of $r$ (where $0 \le r \le 8$). Integers are rational numbers.
• The term $(2)^{8-r}$: Since $2$ is an integer and $8-r$ will also be an integer, $(2)^{8-r}$ will always be an integer, and thus rational.
• The term $(3)^{r/2}$: This is the only part of the term that can introduce irrationality. For $(3)^{r/2}$ to be rational, the exponent $r/2$ must result in an integer. If $r/2$ is not an integer, then $3^{r/2}$ would involve a radical (e.g., $3^{1/2} = \sqrt{3}$, $3^{3/2} = 3\sqrt{3}$), making the entire term irrational.

Therefore, for $T_{r+1}$ to be a rational term, the exponent $r/2$ must be an integer. This implies that $r$ must be an even integer.

The index $r$ in the binomial expansion ranges from $0$ to $n$, so in this case, $0 \le r \le 8$. Combining the condition that $r$ must be an even integer with its range, the possible values for $r$ that will yield rational terms are: $r \in \{0, 2, 4, 6, 8\}$ Why this step is taken: We meticulously examine each component of the general term to pinpoint the exact condition required for the term to be rational. The radical part $(\sqrt{3})^r$ is the sole potential source of irrationality, so we focus on its exponent to ensure it simplifies to an integer power of 3.

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3. Step-by-Step Calculation of Rational Terms

Now we will calculate the value of $T_{r+1}$ for each of the identified rational values of $r$:

• For $r=0$ ($T_1$): $T_1 = {}^8 C_0 (2)^{8-0} (3)^{0/2} = 1 \cdot 2^8 \cdot 3^0 = 1 \cdot 256 \cdot 1 = 256$ (${}^8 C_0 = 1$ and $3^0=1$)

• For $r=2$ ($T_3$): $T_3 = {}^8 C_2 (2)^{8-2} (3)^{2/2} = \left(\frac{8 \times 7}{2 \times 1}\right) \cdot 2^6 \cdot 3^1$ $T_3 = 28 \cdot 64 \cdot 3 = 28 \cdot 192 = 5376$

• For $r=4$ ($T_5$): $T_5 = {}^8 C_4 (2)^{8-4} (3)^{4/2} = \left(\frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1}\right) \cdot 2^4 \cdot 3^2$ $T_5 = 70 \cdot 16 \cdot 9 = 70 \cdot 144 = 10080$

• For $r=6$ ($T_7$): We use the property ${}^n C_r = {}^n C_{n-r}$ to simplify calculation: ${}^8 C_6 = {}^8 C_{8-6} = {}^8 C_2$. $T_7 = {}^8 C_6 (2)^{8-6} (3)^{6/2} = {}^8 C_2 (2)^2 (3)^3$ $T_7 = 28 \cdot 4 \cdot 27 = 28 \cdot 108 = 3024$

• For $r=8$ ($T_9$): $T_9 = {}^8 C_8 (2)^{8-8} (3)^{8/2} = 1 \cdot 2^0 \cdot 3^4 = 1 \cdot 1 \cdot 81 = 81$ (${}^8 C_8 = 1$ and $2^0=1$)

Why this step is taken: These calculations provide the numerical values of the specific terms that meet our rationality criterion. Each calculation is a direct application of the general term formula for the identified values of $r$. Accuracy in these calculations is paramount.

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4. Summation of Rational Terms

To find the sum of all rational terms, we simply add the values we calculated for $T_1, T_3, T_5, T_7,$ and $T_9$: $\text{Sum of Rational Terms} = 256 + 5376 + 10080 + 3024 + 81$ $\text{Sum of Rational Terms} = 18817$ Why this step is taken: This is the final step to answer the problem's question. We have identified and calculated all individual rational terms, and now we combine them to get the total sum.

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5. Alternative Method and Important Tips

Alternative Method (Conjugate Pair Property): For an expansion of the form $(x+\sqrt{y})^n$, where $x$ and $y$ are rational and $n$ is an integer, the sum of rational terms can be found using a clever identity involving its conjugate: Let $S = (x+\sqrt{y})^n$ and $S' = (x-\sqrt{y})^n$. The expansion of $S$ will have terms of the form ${}^n C_r x^{n-r} (\sqrt{y})^r$. The expansion of $S'$ will have terms of the form ${}^n C_r x^{n-r} (-\sqrt{y})^r = {}^n C_r x^{n-r} (-1)^r (\sqrt{y})^r$.

When we add $S$ and $S'$: $S + S' = \sum_{r=0}^{n} {}^n C_r x^{n-r} (\sqrt{y})^r + \sum_{r=0}^{n} {}^n C_r x^{n-r} (-1)^r (\sqrt{y})^r$ $S + S' = \sum_{r=0}^{n} {}^n C_r x^{n-r} (\sqrt{y})^r (1 + (-1)^r)$

• If $r$ is even, $1 + (-1)^r = 1 + 1 = 2$. These are the rational terms.
• If $r$ is odd, $1 + (-1)^r = 1 - 1 = 0$. These are the irrational terms.

Thus, $S + S' = 2 \times (\text{Sum of rational terms})$. So, $\text{Sum of rational terms} = \frac{1}{2} \left[ (x+\sqrt{y})^n + (x-\sqrt{y})^n \right]$.

In our case, $x=2$, $y=3$, $n=8$: Sum of rational terms $= \frac{1}{2} \left[ (2+\sqrt{3})^8 + (2-\sqrt{3})^8 \right]$. Notice that $(2-\sqrt{3})$ is the conjugate of $(2+\sqrt{3})$. Also, $2^2 - (\sqrt{3})^2 = 4 - 3 = 1$. This means $2-\sqrt{3} = \frac{1}{2+\sqrt{3}}$. So, the expression can be written as: Sum of rational terms $= \frac{1}{2} \left[ (2+\sqrt{3})^8 + \left(\frac{1}{2+\sqrt{3}}\right)^8 \right]$. This method provides a powerful shortcut, especially in competitive exams, and can be used to quickly verify results. Calculating $(2+\sqrt{3})^8$ and $(2-\sqrt{3})^8$ directly might be tedious, but if we had already performed the expansion, this identity would confirm