Respuesta :
Answer:
[tex](\sqrt{3} + \frac{1}{\sqrt[4]{6}})^{15}[/tex]
Binomial expansion formula,
[tex](a+b)^n=\sum_{r=0}^{n} ^nC_r (a)^{n-r} (b)^r[/tex]
Where,
[tex]^nC_r=\frac{n!}{r!(n-r)!}[/tex]
[tex]\implies (\sqrt{3} + \frac{1}{2})^{15}=\sum_{r=0}^{15} ^{15}C_r (\sqrt{3})^{15-r} (\frac{1}{\sqrt[4]{6}})^r[/tex]
[tex]=(\sqrt{3})^{15}+15(\sqrt{3})^{14}(\frac{1}{\sqrt[4]{6}})^1+105(\sqrt{3})^{13}(\frac{1}{\sqrt[4]{6}})^2+455(\sqrt{3})^{12}(\frac{1}{\sqrt[4]{6}})^3+1365(\sqrt{3})^{11}(\frac{1}{\sqrt[4]{6}})^4+3003(\sqrt{3})^{10}(\frac{1}{\sqrt[4]{6}})^5+5005(\sqrt{3})^{9}(\frac{1}{\sqrt[4]{6}})^6+6435(\sqrt{3})^{8}(\frac{1}{\sqrt[4]{6}})^7+6435(\sqrt{3})^{7}(\frac{1}{\sqrt[4]{6}})^8+5005(\sqrt{3})^{6}(\frac{1}{\sqrt[4]{6}})^9+3003(\sqrt{3})^{5}(\frac{1}{\sqrt[4]{6}})^{10}+1365(\sqrt{3})^{4}(\frac{1}{\sqrt[4]{6}})^{11}+455(\sqrt{3})^{3}(\frac{1}{\sqrt[4]{6}})^{12}+105(\sqrt{3})^{2}(\frac{1}{\sqrt[4]{6}})^{13}+15(\sqrt{3})^{1}(\frac{1}{\sqrt[4]{6}})^{14}+(\frac{1}{\sqrt[4]{6}})^{15}[/tex]
∵ both [tex]\sqrt{3}[/tex] and [tex]\frac{1}{\sqrt[4]{6}}[/tex] are irrational numbers,
And, if the power of √3 is even, it converted to a rational number,
If its power is odd it remained as irrational number,
But, the product of a rational number and irrational number is irrational,
Thus, all terms in the above expansion are irrational. ( which can not expressed in the form of p/q, where, p and q are integers s.t. q ≠ 0 )
