Question

The sum of the 24 terms of the following series \[\sqrt 2 + \sqrt 8 + \sqrt {18} + \sqrt {32} \] ….is
A. 300
B. \[200\sqrt 2 \]
C. \[300\sqrt 2 \]
D. \[250\sqrt 2 \]

Answer 1

Hint: If we observe the series so given the roots can be expressed as multiples of 2. Thus we will write them in that form. Then we will conclude that there are perfect squares under the root. We will take them outside in their root form. After that the root 2 will be taken common leaving behind a sum of series of consecutive natural numbers. And we know how to find the sum of consecutive natural numbers.

Complete step by step answer:
Given the series is,
\[\sqrt 2 + \sqrt 8 + \sqrt {18} + \sqrt {32} \]
Now this can be written as,
\[ = \sqrt 2 + \sqrt {4 \times 2} + \sqrt {9 \times 2} + \sqrt {16 \times 2} ....\]
Take out the perfect squares out of the root,
\[ = \sqrt 2 + 2\sqrt 2 + 3\sqrt 2 + 4\sqrt 2 ...\]
Now take root 2 common,
\[ = \sqrt 2 \left( {1 + 2 + 3 + 4 + ...} \right)\]
Now the bracket consists of the sum of consecutive natural numbers, and it is given by \[\dfrac{{n\left( {n + 1} \right)}}{2}\]
Here we have n=24 terms. thus putting that value,
\[
   = \sqrt 2 \left( {\dfrac{{24\left( {24 + 1} \right)}}{2}} \right) \\
   = \sqrt 2 \left( {\dfrac{{24 \times 25}}{2}} \right) \\
   = \sqrt 2 \left( {12 \times 25} \right) \\
   = \sqrt 2 \times 300 \\
 \]
On rearranging the terms we get.
\[ = 300\sqrt 2 \]

So, the correct answer is “Option C”.

Note: This is a very basic but logical question. Even the options can help us in proceeding towards the solution sometimes. We simply need to find the clue. The formula used is applicable only because the series has consecutive terms, otherwise we have to find the logic or pattern of the series and then decide which formula of sum is to be used.