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

Question

The sum of all rational terms in the expansion of $(2+\sqrt{3}{)}^{8}$ is :

Accepted Answer

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

$S={}^{8}{C}_{0}(2{)}^{8}+{}^{8}{C}_{1}(2{)}^{7}(\sqrt{3})+\ldots +{}^{8}{C}_{8}(\sqrt{3}{)}^{8}$

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:

$\begin{gathered} {}^{8}{C}_{0}(2{)}^{8}+{}^{8}{C}_{2}(2{)}^{6}(\sqrt{3}{)}^{2}+{}^{8}{C}_{4}(2{)}^{4}(\sqrt{3}{)}^{4}+ \ {}^{8}{C}_{6}(2{)}^{2}(\sqrt{3}{)}^{6}+{}^{8}{C}_{8}(\sqrt{3}{)}^{8} \end{gathered}$

Evaluating these terms:

${}^{8}{C}_{0}(2{)}^{8}=256$

${}^{8}{C}_{2}(2{)}^{6}(3)={}^{8}{C}_{2}	imes 64	imes 3=1344$

${}^{8}{C}_{4}(2{)}^{4}(3{)}^{2}={}^{8}{C}_{4}	imes 16	imes 9=3024$

${}^{8}{C}_{6}(2{)}^{2}(3{)}^{3}={}^{8}{C}_{6}	imes 4	imes 27=4032$

${}^{8}{C}_{8}(3{)}^{4}=1	imes 81=81$

Summing these rational terms gives:

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

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

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