Question

If $\sqrt{2}+\sqrt{6}+\sqrt{18}+......$ 10 terms is equal to $k(\sqrt{2}+\sqrt{6})$, then find the value of k.

Answer 1

$\sqrt{2}+\sqrt{6}+\sqrt{18}+\cdots 10terms=\sqrt{2}(1+\sqrt{3}+\sqrt{9}+\cdots )=\sqrt{2}(\frac{1\left(\right(\sqrt{3}{)}^{10}-1}{\sqrt{3}-1})=\sqrt{2}\times \frac{\sqrt{3}+1}{3-1}\times (3^5-1)=121(\sqrt{2}+\sqrt{6})$

$⟹k(\sqrt{2}+\sqrt{6})=121(\sqrt{2}+\sqrt{6})⟹k=121$