Problem 54
Question
The sum of first \(n\) terms of the series \(1 \cdot 1 !+2 \cdot 2 !+3 \cdot 3 !+4 \cdot 4 !+\ldots\) is (A) \((n+1) !-1\) (B) \(n !-1\) (C) \((n-1) !-1\) (D) None of these
Step-by-Step Solution
Verified Answer
The answer is (A) \((n+1)! - 1\).
1Step 1: Understanding the Series
The given series is a sum of terms of the form \( k \cdot k! \), where \( k \) ranges from 1 to \( n \). We write this sum as \( \sum_{k=1}^{n} k \cdot k! \).
2Step 2: Identifying the Pattern
Observe that \( k \cdot k! = (k+1)! - k! \). This can be seen by expanding \( (k+1)! = (k+1) \cdot k! = k! + k \cdot k! \). Thus, \( k \cdot k! = (k+1)! - k! \).
3Step 3: Rewriting the Series
Using the identity from the previous step, the series \( \sum_{k=1}^{n} k \cdot k! \) can be rewritten as a telescoping sum: \( \sum_{k=1}^{n} ((k+1)! - k!) \). As a result, many terms cancel out.
4Step 4: Evaluating the Telescopic Sum
In a telescopic series of the form \( \sum_{k=1}^{n} ((k+1)! - k!) \), the intermediate terms cancel, leaving only the last term of the first expression and the initial term of the last expression. Therefore, we get \( (n+1)! - 1! = (n+1)! - 1 \).
5Step 5: Choosing the Correct Option
From the evaluated final term of the sum, we determine that the series simplifies to \( (n+1)! - 1 \). Therefore, the correct answer is (A) \((n+1)! - 1\).
Key Concepts
Telescoping SeriesFactorialsMathematical Induction
Telescoping Series
A telescoping series is a special type of series in mathematics where most of the terms cancel each other out when the series is expanded. The magic of a telescoping series lies in its ability to simplify complex summations by eliminating terms through cancellation. This happens because each term often contains elements that negate parts of the previous or following terms.
For example, in the given series:
This is why we call it telescoping—it effectively "collapses" to reveal only the first and last elements of the sequence: \( (n+1)! - 1 \). This feature makes telescoping series very useful in simplifying and evaluating complex sums efficiently.
For example, in the given series:
- We have each term of the form \( (k+1)! - k! \).
- When expanded, the series looks like \( ((2)! - 1!) + ((3)! - 2!) + ((4)! - 3!) + ... + ((n+1)! - n!) \).
- After expansion, nearly all terms cancel out with their neighbors.
This is why we call it telescoping—it effectively "collapses" to reveal only the first and last elements of the sequence: \( (n+1)! - 1 \). This feature makes telescoping series very useful in simplifying and evaluating complex sums efficiently.
Factorials
Factorials are a fundamental concept in mathematics, mainly involved in permutations and combinations, as well as series and sequences. A factorial, denoted by an exclamation mark (!), is the product of all positive integers up to a specified number.
For example:
Factorials grow very quickly in size as \( n \) increases, which makes them interesting and vital in various mathematical problems.In the telescoping series from the exercise, factorials help facilitate the cancellation of terms.
Each term \( k \cdot k! \) simplifies when rewritten using factorials: \( (k+1)! - k! \). This identity is crucial, as it forms the basis for the telescoping property, allowing the series to be solved more easily.
For example:
- \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \)
- \( 3! = 3 \times 2 \times 1 = 6 \)
Factorials grow very quickly in size as \( n \) increases, which makes them interesting and vital in various mathematical problems.In the telescoping series from the exercise, factorials help facilitate the cancellation of terms.
Each term \( k \cdot k! \) simplifies when rewritten using factorials: \( (k+1)! - k! \). This identity is crucial, as it forms the basis for the telescoping property, allowing the series to be solved more easily.
Mathematical Induction
Mathematical induction is a powerful proof technique used to establish the validity of a statement for all natural numbers. It is a bit like building a staircase: if you prove you can reach the first step, and can always move from one step to the next, you can climb indefinitely.
The process generally involves two main steps:
Although mathematical induction was not explicitly used in the original problem, its principles are often applied in sequences and series, especially when proving the validity of formulas that generalize to all natural numbers. It ensures that a formula or expression, such as the simplified sum we arrived at \( (n+1)! - 1 \), holds true for all values in its domain. This builds trust in the generalization beyond just computation for isolated cases.
The process generally involves two main steps:
- **Base Case:** Verify that the statement holds for the first natural number, typically \( n = 1 \).
- **Inductive Step:** Assume the statement is true for some arbitrary natural number \( n = k \), and then prove it holds for \( n = k+1 \).
Although mathematical induction was not explicitly used in the original problem, its principles are often applied in sequences and series, especially when proving the validity of formulas that generalize to all natural numbers. It ensures that a formula or expression, such as the simplified sum we arrived at \( (n+1)! - 1 \), holds true for all values in its domain. This builds trust in the generalization beyond just computation for isolated cases.
Other exercises in this chapter
Problem 52
The sum of \(n\) terms of \(m\) A.P.s are \(S_{1}, S_{2}, S_{3}, \ldots, S_{m}\), If the first term and common difference are \(1,2,3, \ldots, m\) respectively,
View solution Problem 53
If three positive numbers \(a, b, c\) are in H.P., then \(a^{n}+c^{n}\) \((\mathrm{A})>2 b^{n}\) \((\mathrm{B})=2 b^{n}\) \((\mathrm{C})b^{n}\)
View solution Problem 55
If \(a, b, c\) are digits, then the rational number represented by \(0 \cdot c a b a b a b \ldots\) is (A) \(\frac{99 c+a b}{990}\) (B) \(\frac{99 c+10 a+b}{99}
View solution Problem 56
The sum of first \(n\) terms of the series \(1^{2}+2.2^{2}+3^{2}+2.4^{2}+5^{2}+5.6^{2}+\ldots\) is \(\frac{n(n+1)^{2}}{2}\) when \(n\) is even. When \(n\) is od
View solution