Problem 26
Question
\(S=\tan ^{-1}\left(\frac{1}{n^{2}+n+1}\right)+\tan ^{-1}\left(\frac{1}{n^{2}+3 n+3}\right)+\ldots\) \(+\tan ^{-1}\left(\frac{1}{1+(n+19)(n+20)}\right)\), then \(\tan S\) is equal to: (a) \(\frac{20}{401+20 n}\) (b) \(\frac{n}{n^{2}+20 n+1}\) (c) \(\frac{20}{n^{2}+20 n+1}\) (d) \(\frac{n}{401+20 n}\)
Step-by-Step Solution
Verified Answer
The correct answer is (c): \( \frac{20}{n^2 + 20n + 1} \).
1Step 1: Understanding the Pattern
The given sum involves the function \( \tan^{-1} \left(\frac{1}{k^2+k+1}\right) \) going from \( n^2+n+1 \) to \( (n+19)(n+20)+1 \), for values of \( k = n \) to \( n+19 \). This suggests a telescoping nature of the series.
2Step 2: Recognizing Telescoping Series
Notice that \( \tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1} \left( \frac{x+y}{1-xy} \right) \). Using this identity helps simplify the series as a telescoping series, allowing terms to cancel.
3Step 3: Applying the Telescoping Property
Each term fits the identity \( \tan^{-1} x + \tan^{-1} y = \tan^{-1} \left( \frac{x+y}{1-xy} \right) \). By comparing consecutive terms, we group pairs. Pairs cancel each other out except the first term of the first component and the last term of the last component.
4Step 4: Calculating the Boundary Values
For the boundary of this telescoping series, we check the values at \( k = n \) and \( k = n+19 \). As the expression is telescoping, most terms cancel, leaving the result dependent only on the limits.
5Step 5: Simplifying the Result
Simplifying the remaining uncanceled parts, we arrive at a fraction in terms of \( n \) that results from the telescoping elimination. Substituting into the identity gives us the final expression.
6Step 6: Final Calculation
After simplification and applying the tangent sum identity, the expression becomes simplified to match one of the given options, specifically (c): \( \frac{20}{n^2 + 20n + 1} \).
Key Concepts
Telescoping SeriesTrigonometric IdentitiesMathematical Series Simplification
Telescoping Series
A telescoping series is a special type of mathematical series where most terms cancel out, yielding a simplified result. This is a powerful method in summing series as it reduces the complexity of calculations.
Here's how it works: In a telescoping series, consecutive terms are designed such that they "collapse" into just a few terms, usually reducing to the first and last terms of the series. For example, in our exercise, we have the expression \( \tan^{-1} \left(\frac{1}{k^2 + k + 1}\right)\)
- As you pair terms, each pair cancels a portion of the previous terms.- The telescoping property is evident when the terms subtract or "telescope" into a more straightforward expression.- This leaves the boundary terms, which are those at the start or end of the series, responsible for the final result.
Telescoping effectively allows one to compute the sum of a long series with minimal effort, due to the cancellation of intermediate terms.
Here's how it works: In a telescoping series, consecutive terms are designed such that they "collapse" into just a few terms, usually reducing to the first and last terms of the series. For example, in our exercise, we have the expression \( \tan^{-1} \left(\frac{1}{k^2 + k + 1}\right)\)
- As you pair terms, each pair cancels a portion of the previous terms.- The telescoping property is evident when the terms subtract or "telescope" into a more straightforward expression.- This leaves the boundary terms, which are those at the start or end of the series, responsible for the final result.
Telescoping effectively allows one to compute the sum of a long series with minimal effort, due to the cancellation of intermediate terms.
Trigonometric Identities
Trigonometric identities are equations involving trigonometric functions that are true for every value of the occurring variables. They are essential when working with trigonometric expressions because they enable the simplification of complex problems.
One critical identity used in the original exercise involves the tangent function:\( \tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1} \left( \frac{x+y}{1-xy} \right) \).
One critical identity used in the original exercise involves the tangent function:\( \tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1} \left( \frac{x+y}{1-xy} \right) \).
- This particular identity allows you to combine angles, making simplification more manageable.
- It is especially helpful in telescoping series because it explains how two inverse tangents can be grouped into a single term.
- Recognizing when and how to apply these identities is key in simplifying and solving inverse trigonometry problems effectively.
Mathematical Series Simplification
Mathematical series simplification is the process of reducing a complex series into a simpler form or sum. This often involves techniques that make calculations more efficient, particularly for long series.
There are several strategies to simplify a series:
In the given exercise, these methods are fundamental. They simplify a potentially cumbersome task into a more manageable one, often leading directly to an answer by minimizing the need for computation beyond the key terms. It's these simplification strategies that can reduce a lengthy expression to a neat fraction, directly answering the problem posed. This approach exemplifies the power and elegance of mathematical solutions.
There are several strategies to simplify a series:
- **Grouping:** Like terms can often be grouped to cancel each other out, as we see in telescoping series.
- **Trigonometric Identities:** Utilizing identities, particularly tangent identities, can simplify trigonometric series problems.
- **Boundary Analysis:** Focusing on starting and ending terms, since they typically remain after cancellation.
In the given exercise, these methods are fundamental. They simplify a potentially cumbersome task into a more manageable one, often leading directly to an answer by minimizing the need for computation beyond the key terms. It's these simplification strategies that can reduce a lengthy expression to a neat fraction, directly answering the problem posed. This approach exemplifies the power and elegance of mathematical solutions.
Other exercises in this chapter
Problem 24
If \(x, y, z\) are in A.P. and \(\tan ^{-1} x, \tan ^{-1} y\) and \(\tan ^{-1} z\) are also in A.P., then (a) \(x=y=z\) (b) \(2 x=3 y=6 z\) (c) \(6 x=3 y=2 z\)
View solution Problem 25
Let \(x \in(0,1) .\) The set of all \(x\) such that \(\sin ^{-1} x>\cos ^{-1} x\), is the interval: \([\) Online April \(25,2013 \mid\) (a) \(\left(\frac{1}{2},
View solution Problem 27
A value of \(x\) for which \(\sin \left(\cot ^{-1}(1+x)\right)=\cos \left(\tan ^{-1} x\right)\), is (a) \(-\frac{1}{2}\) (b) 1 (c) 0 (d) \(\frac{1}{2}\)
View solution Problem 28
If \(\sin ^{-1}\left(\frac{x}{5}\right)+\operatorname{cosec}^{-1}\left(\frac{5}{4}\right)=\frac{\pi}{2}\), then the values of \(\mathrm{x}\) is (a) 4 (b) \(5 \)
View solution