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 = $$2\sqrt 2 $$ $$ - $$ $$\sqrt 2 $$ = $$\sqrt 2 $$
$$\eqalign{ & \therefore {S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right] \cr & = \frac{n}{2}\left[ {2\sqrt 2 + \left( {n - 1} \right)\sqrt 2 } \right] \cr & = \frac{n}{2}\left[ {2\sqrt 2 + \sqrt 2 \,n - \sqrt 2 } \right] \cr & = \frac{n}{2}\left( {\sqrt 2 \,n + \sqrt 2 } \right) \cr & = \frac{{n\sqrt 2 }}{2}\left( {n + 1} \right) \cr & = \frac{{n\left( {n + 1} \right)}}{{\sqrt 2 }} \cr} $$
This Question Belongs to Arithmetic Ability >> Progressions
Step 1: Simplify the series first.
The series looks complicated, but it's not. Let's simplify each square root:
√2 = 1√2
√8 = √(4 × 2) = 2√2
√18 = √(9 × 2) = 3√2
√32 = √(16 × 2) = 4√2
So the series is actually: 1√2 + 2√2 + 3√2 + 4√2 + ...
Step 2: Recognize the pattern.
This is a simple Arithmetic Progression (A.P.).
The first term (a) is √2.
The common difference (d) is also √2 (since each term increases by √2).
Step 3: Use the sum formula for an A.P.
The formula is: Sₙ = n/2 * [2a + (n-1)d]
Now, just plug in our values for a and d:
Sₙ = n/2 * [2(√2) + (n-1)√2]
Step 4: Simplify the expression.
Factor out √2 from inside the bracket:
Sₙ = n/2 * [ √2 * (2 + n - 1) ]
Simplify inside the parenthesis:
Sₙ = n/2 * [ √2 * (n + 1) ]
Rearrange it:
Sₙ = ( n * (n + 1) * √2 ) / 2
Since 2 = √2 * √2, we can cancel one √2 from the top and bottom:
Sₙ = n(n + 1) / √2
This matches the answer.
Therefore, the correct option is C.