Question

Sum of rational terms in $(\sqrt{2}+\sqrt[5]{53}{)}^{10}$ is

• A. $41$
• B. $42$
• C. $32$
• D. $38$

Answer 1

The correct option is A $41$

$(a+b{)}^n={\sum }_{k=0}^n{a}^{n-k}{b}^k$
Now, ${(\sqrt{2}+\sqrt[5]{53})}^{10}={\sum }_{k=0}^n{\sqrt{2}}^{n-k}{\sqrt[5]{53}}^k$
We get rational values only for $k=0,10$
Thus Sum of rational numbers is $(\frac{10}{0}){\sqrt{2}}^{10-0}{\sqrt[5]{53}}^0+(\frac{10}{10}){\sqrt{2}}^{10-10}{\sqrt[5]{53}}^{10}$

$\frac{10!}{0!(10-0)!}{2}^5+\frac{10!}{10!(10-0)!}{3}^2$

$\frac{10!}{10!}{2}^5+\frac{10!}{10!}{3}^2$

$32+9=41$