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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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