# Sum of n terms of the series $$\sqrt 2$$  $$+$$  $$\sqrt 8$$  $$+$$  $$\sqrt {18}$$  $$+$$  $$\sqrt {32}$$  $$+$$  ....... is

A. $$\frac{{n\left( {n + 1} \right)}}{2}$$

B. $$2n\left( {n + 1} \right)$$

C. $$\frac{{n\left( {n + 1} \right)}}{{\sqrt 2 }}$$

D. $$1$$

Answer: Option C

### Solution(By Examveda Team)

The series is given
$$\sqrt 2 + \sqrt 8 + \sqrt {18} + \sqrt {32}$$     $$+$$ ......
$$\Rightarrow \sqrt 2 + 2\sqrt 2 + 3\sqrt 2$$     $$+$$ $$4\sqrt 2$$  $$+$$ ......
Here a = $$\sqrt 2$$   and d = $$2\sqrt 2$$  $$-$$ $$\sqrt 2$$  = $$\sqrt 2$$
\eqalign{ & \therefore {S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right] \cr & = \frac{n}{2}\left[ {2\sqrt 2 + \left( {n - 1} \right)\sqrt 2 } \right] \cr & = \frac{n}{2}\left[ {2\sqrt 2 + \sqrt 2 \,n - \sqrt 2 } \right] \cr & = \frac{n}{2}\left( {\sqrt 2 \,n + \sqrt 2 } \right) \cr & = \frac{{n\sqrt 2 }}{2}\left( {n + 1} \right) \cr & = \frac{{n\left( {n + 1} \right)}}{{\sqrt 2 }} \cr}
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