Question

The sum to \(n\) terms of the series \(\frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{7}}+\ldots\) is : (a) \(\sqrt{2 n+1}\) (b) \(\frac{1}{2} \sqrt{2 n+1}\) (c) \(\sqrt{2 n-1}\) (d) \(\frac{1}{2}\\{\sqrt{2 n+1}-1\\}\)

Short answer

Question: Find the sum of the series \(\frac{1}{\sqrt{2k-1}+\sqrt{2k+1}}\) for k=1 to k=n. Answer: \(\frac{1}{2}(\sqrt{2n+1}-\sqrt{1})\)

Step-by-step solution

Identify the pattern

We can rewrite each term of the series using the identity \(\frac{a}{a+b}+\frac{b}{a+b}=1\), which lets us write the series as: $$\frac{1}{\sqrt{1}+\sqrt{3}}=\frac{\sqrt{3}-\sqrt{1}}{(\sqrt{1}+\sqrt{3})(\sqrt{3}-\sqrt{1})}=\frac{\sqrt{3}-\sqrt{1}}{2}$$ Similarly: $$\frac{1}{\sqrt{3}+\sqrt{5}}=\frac{\sqrt{5}-\sqrt{3}}{(\sqrt{3}+\sqrt{5})(\sqrt{5}-\sqrt{3})}=\frac{\sqrt{5}-\sqrt{3}}{2}$$

Rewrite the series

Now, rewrite the series with the simplified terms: $$\frac{1}{2}(\sqrt{3}-\sqrt{1})+\frac{1}{2}(\sqrt{5}-\sqrt{3})+\frac{1}{2}(\sqrt{7}-\sqrt{5})+\ldots$$

Find the sum to n terms

We can see that there are many terms that cancel each other out. So the sum of the series to n terms can be obtained by cancelling the common terms, leaving us with: $$\frac{1}{2}(\sqrt{2n+1}-\sqrt{1}).$$

Compare with the given options

Comparing our result with the given options, we see that our answer matches option (d): $$ \frac{1}{2}\\{\sqrt{2 n+1}-1\\} $$

Key concepts

Arithmetic Progression

An arithmetic progression (AP) is a sequence of numbers in which each term after the first is obtained by adding a constant, known as the common difference, to the previous term. For instance, if we start with the number 2 and keep adding 3, we get the series 2, 5, 8, 11, which is an example of an AP where the common difference is 3.

The key to mastering problems involving AP is recognizing the pattern and applying the formula for the sum of the first n terms which is given by: \[ S_n = \frac{n}{2}(2a + (n-1)d) \], where \( S_n \) is the sum of the first n terms, \( a \) is the first term, \( n \) is the number of terms, and \( d \) is the common difference.

Sequence and Series

In mathematics, sequences and series are closely related concepts. A sequence is an ordered list of numbers following a specific rule, while a series is the sum of the terms of a sequence.

For instance, the sequence 1, 4, 7, 10, ... continues indefinitely according to a simple pattern: starting with 1, each term adds 3. If we sum these terms, we get a series: 1 + 4 + 7 + 10 + ... This can also be further classified into different types such as arithmetic series, geometric series, and so on, each with its own unique properties and formulas for calculation.

When approaching series problems, it's crucial to first discern the type of sequence you're dealing with—whether it's arithmetic, geometric, or something else altogether. This distinction dictates the strategy for finding the sum.

Mathematical Induction

Mathematical induction is a powerful proof technique used to prove the validity of a statement for all natural numbers. It consists of two main steps. First is the 'base case', where the statement is verified for the initial value, typically \(n = 1\). Next is the 'induction step', where one assumes the statement holds for some arbitrary natural number \(n\) and then proves it for \(n + 1\).

This is akin to proving that a series of dominos will fall: if we know the first domino falls, and can prove that every time one domino falls it will cause the next one to fall, then we can conclude all the dominos will fall. Mathematical induction is a form of direct proof, as opposed to proof by contradiction or proof by contrapositive.

Algebraic Simplification

Algebraic simplification is the process of reducing expressions to their simplest form using a variety of algebraic rules and techniques such as distributing, factoring, combining like terms, and canceling.

In the given problem, algebraic simplification is used to rewrite the terms in the series in a form that reveals the cancellation amongst them. This makes it possible to find a pattern or simpler expression, especially in complex series or sequence problems. Such simplifications often involve recognizing and applying algebraic identities, as seen with the identity \(\frac{a}{a+b} + \frac{b}{a+b} = 1\).

By simplifying each term, one can often disclose the underlying structure of the series, which in this case, led to identifying the cancellations and finding the sums up to the n-th term with relative ease.