Chapter 1: Problem 44
\(\frac{1}{\sqrt{9}+\sqrt{10}}+\frac{1}{\sqrt{10}+\sqrt{11}}+\frac{1}{\sqrt{11}+\sqrt{12}} \cdots \cdots \quad\) up to 91 terms is \begin{tabular}{l} \hline \end{tabular} (1) 7 (2) 8 (3) 6 (4) 9
Short Answer
Expert verified
Solution: The sum of the first 91 terms in the sequence is 7.
Step by step solution
01
Identify the pattern
Looking at the given series, we can see that each term can be represented as follows:
\[\frac{1}{\sqrt{n}+\sqrt{n+1}}\]
Where n starts from 9 and we are interested in the sum of the first 91 terms.
02
Rationalize the denominator
In order to handle the square roots, we can rationalize the denominator for a general term. We can do this by multiplying the numerator and the denominator by a conjugate of the denominator:
\[\frac{1}{\sqrt{n}+\sqrt{n+1}} \cdot \frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n+1}-\sqrt{n}}\]
03
Simplify the general term
Now, we can simplify the expression to find the general term:
\[\frac{\sqrt{n+1}-\sqrt{n}}{((\sqrt{n+1})^2 - (\sqrt{n})^2)}\]
\[\frac{\sqrt{n+1}-\sqrt{n}}{n+1-n}\]
\[\frac{\sqrt{n+1}-\sqrt{n}}{1}\]
So, our general term is:
\[T_n = \sqrt{n+1}-\sqrt{n}\]
04
Sum the first 91 terms
Now, we want to find the sum of the first 91 terms, which can be represented as:
\[S_{91} = T_9 + T_{10} + T_{11} ... + T_{99}\]
We know the relationship between the terms:
\[T_n = \sqrt{n+1}-\sqrt{n}\]
Therefore, we can substitute the relationship into the sum expression:
\[S_{91} = (\sqrt{10}-\sqrt{9}) + (\sqrt{11}-\sqrt{10}) + (\sqrt{12}-\sqrt{11}) + ... + (\sqrt{100}-\sqrt{99})\]
05
Simplify the sum expression
Looking at the expression, we can see that most terms will cancel out, leaving only the first and the last terms:
\[S_{91} = \sqrt{100} - \sqrt{9}\]
06
Calculate the final result
We can evaluate the final expression:
\[S_{91} = 10 - 3\]
\[S_{91} = 7\]
So, the sum of the first 91 terms is 7, which corresponds to option (1).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Rationalizing the Denominator
The process of rationalizing the denominator is used when you have a fraction with a radical (usually a square root) in the denominator.
The goal of rationalizing is to eliminate the radical from the denominator, which not only simplifies the expression but also provides a form that is generally more useful in further operations and less ambiguous for further numerical approximations.
To rationalize a denominator that contains a square root, you multiply the numerator and the denominator of the fraction by the conjugate of the denominator. The conjugate of \(\sqrt{a} + \sqrt{b}\) is \(\sqrt{a} - \sqrt{b}\). When we multiply these conjugates, the result under the radical sign is the difference of two squares, which simplifies the radicals completely.
Here is a clear example:
\[\frac{1}{\sqrt{a}+\sqrt{b}} \times \frac{\sqrt{b}-\sqrt{a}}{\sqrt{b}-\sqrt{a}}\]
This simplifies to:
\[\frac{\sqrt{b}-\sqrt{a}}{b-a}\]
Using this method, it makes the original complex fractions much simpler to work with and is crucial for solving various mathematical problems.
The goal of rationalizing is to eliminate the radical from the denominator, which not only simplifies the expression but also provides a form that is generally more useful in further operations and less ambiguous for further numerical approximations.
To rationalize a denominator that contains a square root, you multiply the numerator and the denominator of the fraction by the conjugate of the denominator. The conjugate of \(\sqrt{a} + \sqrt{b}\) is \(\sqrt{a} - \sqrt{b}\). When we multiply these conjugates, the result under the radical sign is the difference of two squares, which simplifies the radicals completely.
Here is a clear example:
\[\frac{1}{\sqrt{a}+\sqrt{b}} \times \frac{\sqrt{b}-\sqrt{a}}{\sqrt{b}-\sqrt{a}}\]
This simplifies to:
\[\frac{\sqrt{b}-\sqrt{a}}{b-a}\]
Using this method, it makes the original complex fractions much simpler to work with and is crucial for solving various mathematical problems.
Sequences and Series
In mathematics, sequences and series are closely related concepts that involve ordered lists of numbers.
A sequence is a set of numbers listed in a specific order, following a certain rule. For example, the sequence of squares is 1, 4, 9, 16, ..., where each term is the square of an integer.
A series is the sum of the terms of a sequence. A finite series has a last term, while an infinite one continues indefinitely. The series corresponding to the sequence of squares is 1 + 4 + 9 + 16 + ....
To analyze series, mathematicians often look for patterns in the sequence of terms and employ various techniques to find the sum. As in our exercise, we can look for cancellation or telescoping to simplify the series to its fundamental components, which aids in finding the sum.
A sequence is a set of numbers listed in a specific order, following a certain rule. For example, the sequence of squares is 1, 4, 9, 16, ..., where each term is the square of an integer.
A series is the sum of the terms of a sequence. A finite series has a last term, while an infinite one continues indefinitely. The series corresponding to the sequence of squares is 1 + 4 + 9 + 16 + ....
To analyze series, mathematicians often look for patterns in the sequence of terms and employ various techniques to find the sum. As in our exercise, we can look for cancellation or telescoping to simplify the series to its fundamental components, which aids in finding the sum.
Square Root Simplification
The concept of square root simplification is essential whenever you’re dealing with square roots, especially in complex algebraic expressions.
Simplifying square roots involves finding an equivalent expression where the radical has the simplest possible form. This doesn't always mean removing the radical entirely, but rather expressing the number under the radical as product of a square number and another factor, when possible.
For example:
\[\sqrt{8} = \sqrt{4 \times 2} = \sqrt{4} \times \sqrt{2} = 2\sqrt{2}\]
The expression \(2\sqrt{2}\) is simpler and more useful than \(\sqrt{8}\) because \(2\) and \(\sqrt{2}\) can be more easily manipulated in equations and arithmetic operations.
Simplifying square roots involves finding an equivalent expression where the radical has the simplest possible form. This doesn't always mean removing the radical entirely, but rather expressing the number under the radical as product of a square number and another factor, when possible.
For example:
\[\sqrt{8} = \sqrt{4 \times 2} = \sqrt{4} \times \sqrt{2} = 2\sqrt{2}\]
The expression \(2\sqrt{2}\) is simpler and more useful than \(\sqrt{8}\) because \(2\) and \(\sqrt{2}\) can be more easily manipulated in equations and arithmetic operations.
Arithmetic Progression
An Arithmetic Progression (AP) is one of the most basic types of sequence and series. It is a sequence where the difference between successive terms is constant. This difference is known as the common difference, denoted as \(d\).
An example of an AP is 2, 4, 6, 8, ..., where the common difference \(d=2\). The nth term in an arithmetic progression can be found using the formula:
\[T_n = a + (n-1)d\]
where \(a\) is the first term, and \(n\) is the position of the term in the sequence.
Summing the terms in an AP is vital in various situations, like calculating the total of the first \(n\) natural numbers, interest computations, and statistical applications. The sum of the first \(n\) terms of an AP is given by the formula:
\[S_n = \frac{n}{2}(2a + (n - 1)d)\]
or equivalently, \[S_n = \frac{n}{2}(a + T_n)\],
where \(T_n\) is the nth term. Understanding how arithmetic progression works is key in solving many arithmetic problems, including those found in competitive exams such as the IIT Foundation series.
An example of an AP is 2, 4, 6, 8, ..., where the common difference \(d=2\). The nth term in an arithmetic progression can be found using the formula:
\[T_n = a + (n-1)d\]
where \(a\) is the first term, and \(n\) is the position of the term in the sequence.
Summing the terms in an AP is vital in various situations, like calculating the total of the first \(n\) natural numbers, interest computations, and statistical applications. The sum of the first \(n\) terms of an AP is given by the formula:
\[S_n = \frac{n}{2}(2a + (n - 1)d)\]
or equivalently, \[S_n = \frac{n}{2}(a + T_n)\],
where \(T_n\) is the nth term. Understanding how arithmetic progression works is key in solving many arithmetic problems, including those found in competitive exams such as the IIT Foundation series.
