JEE Main 2025 — Binomial Theorem Question with Solution

To find the sum of all rational terms in the expansion of $(2+\sqrt{3}{)}^8$, we consider the binomial expansion:

$S={}^8{C}_0(2{)}^8+{}^8{C}_1(2{)}^7(\sqrt{3})+\dots +{}^8{C}_8(\sqrt{3}{)}^8$

We need to identify and sum only the rational terms. In the binomial expansion, a term is rational if the exponent of $\sqrt{3}$ is even.

Thus, the rational terms are:

$\begin{array}{rl}& {}^8{C}_0(2{)}^8+{}^8{C}_2(2{)}^6(\sqrt{3}{)}^2+{}^8{C}_4(2{)}^4(\sqrt{3}{)}^4+\\ & {}^8{C}_6(2{)}^2(\sqrt{3}{)}^6+{}^8{C}_8(\sqrt{3}{)}^8\end{array}$

Evaluating these terms:

${}^8{C}_0(2{)}^8=256$

${}^8{C}_2(2{)}^6(3)={}^8{C}_2\times 64\times 3=1344$

${}^8{C}_4(2{)}^4(3{)}^2={}^8{C}_4\times 16\times 9=3024$

${}^8{C}_6(2{)}^2(3{)}^3={}^8{C}_6\times 4\times 27=4032$

${}^8{C}_8(3{)}^4=1\times 81=81$

Summing these rational terms gives:

$256+1344+3024+4032+81=18817$

Therefore, the sum of all rational terms in the expansion is $18,817$.