Question

${10}^{\text{th}}$ term in the expansion 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 ${10}^{\text{th}}$ term of ${(3-\sqrt{\frac{17}{4}+3\sqrt{2}})}^{20}={(\frac{3}{2}-\sqrt{2})}^{20}$
$={}^{20}{C}_9{\left(\frac{3}{2}\right)}^{20-9}(-\sqrt{2}{)}^9$
which is an irrational number.