Question

${10}^{th}$ term of ${(3-\sqrt{\frac{17}{4}+3\sqrt{2}})}^{20}$ is

• A. an irrational number
• B. a rational number
• C. a positive integer
• D. a negative integer

Answer 1

The correct option is A

an irrational number

We have , $\frac{17}{4}+3\sqrt{2}=\frac{1}{4}(9+8+12\sqrt{2})$

= $\frac{1}{4}(3+2\sqrt{2}{)}^2$

$\therefore$ $3-\sqrt{\frac{17}{4}+3\sqrt{2}}=3-\frac{1}{2}(3+2\sqrt{2})$ = $\frac{3}{2}-\sqrt{2}$

Hence, the ${10}^{th}$ term of ${(3-\sqrt{\frac{17}{4}+3\sqrt{2}})}^{20}={(\frac{3}{2}-\sqrt{2})}^{20}$ is ${}^{20}{C}_9{\left(\frac{3}{2}\right)}^{20-9}(\sqrt{2}{)}^9$