The sum of all rational terms in the expansion of (2+√3)⁸ is

The correct answer is Option 2. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q9 – JEE-MAIN 2025

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Home Maths Binomial TheoremQuestion #46

The sum of all rational terms in the expansion of

(2 + \sqrt{3})^{8}

Clue 1 of 3: Binomial Expansion Terms

Recall the general term in the binomial expansion of $(a + b)^{n}$ .

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Ninja Strategy•Magnitude Estimation

Estimate the approximate value of $(2 + \sqrt{3})^{8}$ and use the identity for the sum of rational terms to quickly narrow down the options based on magnitude.

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Step 1: Identify General Term and Rationality Condition✦ Active

The general term in the expansion of $(2 + \sqrt{3})^{8}$ is given by $T_{r + 1} = (\binom{8}{r}) (2)^{8 - r} (\sqrt{3})^{r}$ . For $T_{r + 1}$ to be rational, the power of $\sqrt{3}$ must be an integer. This means $r$ must be an even integer. Possible values for $r$ are $0, 2, 4, 6, 8$ .

Step 2: Apply Sum of Rational Terms Identity○ Expand

The sum of all rational terms in the expansion of $(a + \sqrt{b})^{n}$ is given by $\frac{1}{2} [(a + \sqrt{b})^{n} + (a - \sqrt{b})^{n}]$ . For this problem, $a = 2$ , $b = 3$ , and $n = 8$ . So, the sum of rational terms is $\frac{1}{2} [(2 + \sqrt{3})^{8} + (2 - \sqrt{3})^{8}]$ .

💡 Teacher's Secret Hint

This identity simplifies calculations significantly compared to summing individual terms.

Step 3: Calculate the Sum○ Expand

Let $X = 2 + \sqrt{3}$ and $Y = 2 - \sqrt{3}$ . We need to calculate $\frac{1}{2} (X^{8} + Y^{8})$ . First, calculate $X^{2}$ and $Y^{2}$ : $X^{2} = (2 + \sqrt{3})^{2} = 4 + 3 + 4 \sqrt{3} = 7 + 4 \sqrt{3}$ $Y^{2} = (2 - \sqrt{3})^{2} = 4 + 3 - 4 \sqrt{3} = 7 - 4 \sqrt{3}$ Next, calculate $X^{4}$ and $Y^{4}$ : $X^{4} = (7 + 4 \sqrt{3})^{2} = 49 + 16 \cdot 3 + 2 \cdot 7 \cdot 4 \sqrt{3} = 49 + 48 + 56 \sqrt{3} = 97 + 56 \sqrt{3}$ $Y^{4} = (7 - 4 \sqrt{3})^{2} = 49 + 16 \cdot 3 - 2 \cdot 7 \cdot 4 \sqrt{3} = 49 + 48 - 56 \sqrt{3} = 97 - 56 \sqrt{3}$ Finally, calculate $X^{8}$ and $Y^{8}$ : $X^{8} = (97 + 56 \sqrt{3})^{2} = 97^{2} + (56 \sqrt{3})^{2} + 2 \cdot 97 \cdot 56 \sqrt{3}$ $Y^{8} = (97 - 56 \sqrt{3})^{2} = 97^{2} + (56 \sqrt{3})^{2} - 2 \cdot 97 \cdot 56 \sqrt{3}$ Adding $X^{8}$ and $Y^{8}$ : $X^{8} + Y^{8} = 2 [97^{2} + (56 \sqrt{3})^{2}]$ $97^{2} = 9409$ $(56 \sqrt{3})^{2} = 56^{2} \cdot 3 = 3136 \cdot 3 = 9408$ So, $X^{8} + Y^{8} = 2 [9409 + 9408] = 2 [18817] = 37634$ . The sum of rational terms is $\frac{1}{2} (X^{8} + Y^{8}) = \frac{1}{2} (37634) = 18817$ .

💡 Teacher's Secret Hint

Be careful with squaring terms involving surds. Remember $(a + b)^{2} = a^{2} + b^{2} + 2 a b$ .

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Maths Question 9 – JEE-MAIN 2025

Clue 1 of 3: Binomial Expansion Terms

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