Understanding the Series and Finding the Sum

The given series is \(\sqrt 2 + \sqrt 8 + \sqrt {18} + \sqrt {32} + \ldots\). To find the sum of n terms of this series, we first need to understand the nature of the series. Let's simplify the terms:

• The first term is \(\sqrt 2\).
• The second term is \(\sqrt 8 = \sqrt{4 \times 2} = \sqrt 4 \times \sqrt 2 = 2\sqrt 2\).
• The third term is \(\sqrt {18} = \sqrt{9 \times 2} = \sqrt 9 \times \sqrt 2 = 3\sqrt 2\).
• The fourth term is \(\sqrt {32} = \sqrt{16 \times 2} = \sqrt {16} \times \sqrt 2 = 4\sqrt 2\).

So, the series can be written as \(\sqrt 2, 2\sqrt 2, 3\sqrt 2, 4\sqrt 2, \ldots\). This is an arithmetic progression (AP) because the difference between consecutive terms is constant.

Identifying AP Parameters

In this arithmetic progression:

• The first term, \(a_1\) or \(a\), is \(\sqrt 2\).
• The common difference, \(d\), is the difference between any term and its preceding term. For example, \(a_2 - a_1 = 2\sqrt 2 - \sqrt 2 = \sqrt 2\), and \(a_3 - a_2 = 3\sqrt 2 - 2\sqrt 2 = \sqrt 2\). So, the common difference \(d = \sqrt 2\).

Formula for Sum of n Terms of an AP

The sum of the first n terms of an arithmetic progression is given by the formula:

$$S_n = \frac{n}{2}[2a + (n-1)d]$$

Where:

• \(S_n\) is the sum of the first n terms.
• \(n\) is the number of terms.
• \(a\) is the first term.
• \(d\) is the common difference.

Calculating the Sum of n Terms

Now, substitute the values of \(a = \sqrt 2\) and \(d = \sqrt 2\) into the formula for \(S_n\):

$$S_n = \frac{n}{2}[2(\sqrt 2) + (n-1)\sqrt 2]$$

Simplify the expression inside the square brackets:

$$S_n = \frac{n}{2}[2\sqrt 2 + n\sqrt 2 - \sqrt 2]$$

Combine the like terms inside the brackets:

$$S_n = \frac{n}{2}[(2\sqrt 2 - \sqrt 2) + n\sqrt 2]$$

$$S_n = \frac{n}{2}[\sqrt 2 + n\sqrt 2]$$

Factor out \(\sqrt 2\) from the terms inside the brackets:

$$S_n = \frac{n}{2}[\sqrt 2(1 + n)]$$

Rearrange the terms:

$$S_n = \frac{n(n+1)\sqrt 2}{2}$$

To match the format of the options, we can rewrite the denominator \(2\) as \(\sqrt 2 \times \sqrt 2\):

$$S_n = \frac{n(n+1)\sqrt 2}{\sqrt 2 \times \sqrt 2}$$

Cancel out one \(\sqrt 2\) from the numerator and the denominator:

$$S_n = \frac{n(n+1)}{\sqrt 2}$$

Thus, the sum of the first n terms of the given series is \(\frac{{{\rm{n}}\left( {{\rm{n}} + 1} \right)}}{{\sqrt 2 }}\).

Comparing with Options

Let's compare our derived sum with the given options:

The calculated sum matches option 3.

Revision Table: Sum of AP Series

Additional Information: Simplifying Square Roots

Simplifying square roots of the form \(\sqrt{k \times m}\) where \(k\) is a perfect square is a useful skill for these types of problems. The property used is \(\sqrt{ab} = \sqrt a \times \sqrt b\).

Examples:

• \(\sqrt 8 = \sqrt{4 \times 2} = \sqrt 4 \times \sqrt 2 = 2\sqrt 2\).
• \(\sqrt {18} = \sqrt{9 \times 2} = \sqrt 9 \times \sqrt 2 = 3\sqrt 2\).
• \(\sqrt {32} = \sqrt{16 \times 2} = \sqrt {16} \times \sqrt 2 = 4\sqrt 2\).
• \(\sqrt {50} = \sqrt{25 \times 2} = \sqrt {25} \times \sqrt 2 = 5\sqrt 2\).

Recognizing this pattern helps identify the common difference and the first term in series involving such square roots.