JEE Mathematics 2025 Question with Solution

Given: We need the sum of all rational terms in the expansion of $\left(2+\sqrt{3}\right)^8$.

Find: The required sum and hence the correct option.

A term in the binomial expansion

$$\left(2+\sqrt{3}\right)^8=\sum_{k=0}^{8}\binom{8}{k}2^{8-k}(\sqrt{3})^k$$

is rational only when the exponent of $\sqrt{3}$ is even. So we take $k=0,2,4,6,8$.

Thus the rational terms are

$$\binom{8}{0}(2)^8+\binom{8}{2}(2)^6(\sqrt{3})^2+\binom{8}{4}(2)^4(\sqrt{3})^4+\binom{8}{6}(2)^2(\sqrt{3})^6+\binom{8}{8}(\sqrt{3})^8$$

Now evaluate each term:

$$=2^8+28\cdot 2^6\cdot 3+70\cdot 2^4\cdot 9+28\cdot 2^2\cdot 27+81$$ $$=256+5376+10080+3024+81$$ $$=18817$$

Therefore, the sum of all rational terms is $18817$. Hence, the correct option is D.

Using symmetry, the irrational terms cancel in

$$\left(2+\sqrt{3}\right)^8+\left(2-\sqrt{3}\right)^8$$

So the sum of rational terms is

$$\frac{\left(2+\sqrt{3}\right)^8+\left(2-\sqrt{3}\right)^8}{2}$$

which matches the same value obtained above.

Given: Expression $\left(2+\sqrt{3}\right)^8$.

Find: Sum of only the rational terms.

Use the identity that adding conjugate expansions cancels all irrational terms:

$$\left(2+\sqrt{3}\right)^8+\left(2-\sqrt{3}\right)^8$$

The odd powers of $\sqrt{3}$ cancel, while the even powers remain and get doubled. Therefore,

$$\text{Sum of rational terms}=\frac{\left(2+\sqrt{3}\right)^8+\left(2-\sqrt{3}\right)^8}{2}$$

From the expansion, this equals

$$256+5376+10080+3024+81=18817$$

Therefore, the correct option is D.