[tex]\it a = \dfrac{1}{\sqrt2} + \dfrac{2}{\sqrt8} + \dfrac{3}{\sqrt{18}} + \dfrac{4}{\sqrt{32}} = \dfrac{1}{\sqrt2} \left(1+ \dfrac{2}{\sqrt4} + \dfrac{3}{\sqrt9} + \dfrac{4}{\sqrt{16}} \right )=
\\\;\\ \\\;\\
= \dfrac{1}{\sqrt2} (1+1+1+1) = \dfrac{1}{\sqrt2} \cdot4 =\dfrac{4}{\sqrt2} = \dfrac{4\sqrt2}{2}=2\sqrt2 .[/tex]