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Chapter 3: Problem 10

Prove that \(1(1 !)+2(2 !)+\cdots+n(n !)=(n+1) !-1\), for all \(n \in \mathbb{N}\)

Short Answer

Expert verified
Using mathematical induction, we first verify the base case for \(n = 1\), which results in \(1 = 1\). Next, we assume the given equation holds true for an integer \(k\): \(1(1!)+2(2!)+\cdots+k(k!)=(k+1)!-1\). We then prove the formula for the integer \((k + 1)\) using our inductive hypothesis and find that it also holds true. Hence, we conclude that the given equation is valid for all natural numbers \(n\): \(1(1!)+2(2!)+\cdots+n(n!)=(n+1)!-1\).

Step by step solution

01

Establish the Base Case

To establish the base case, we have to demonstrate the formula's validity when \(n = 1\). The given equation for this condition becomes: \(1(1!) = (1+1)! - 1\) On the left-hand side, we have: \(1(1) = 1\) On the right-hand side, we have: \((1+1)! - 1 = (2)! - 1 = 2 - 1 = 1\) Since both sides are equal (i.e., \(1 = 1\)), the base case holds.
02

Assume the Formula is True for Some Integer k

Now, let us assume the formula holds true for some integer \(k \in \mathbb{N}\); that is: \(1(1!)+2(2!)+\cdots+k(k!)=(k+1)!-1\) This is our inductive hypothesis.
03

Prove the Formula is True for Integer k+1

In this step, we will use our inductive hypothesis to prove the formula for the integer \((k + 1)\). If true, it suggests the formula holds for all natural numbers. We will now consider the summation up to \((k+1)\): \(1(1!)+2(2!)+\cdots+k(k!)+(k+1)((k+1)!)\) Using our inductive hypothesis, we can replace the summation up to \(k\): \((k+1)! - 1 + (k+1)((k+1)!)\) Now, we can factor out \((k + 1)!\): \((k+1)![(k + 1) + 1] - 1\) Simplifying the expression yields: \((k+1)![(k+2)] - 1\) Next, we can recognize the expression \((k+2)\) as \((k + 1) + 1\). So: \((k+1)!((k+1)+1) - 1\) This equation exactly matches our formula for the next integer \(k + 1\): \((k+2)! - 1\) Since we have now demonstrated that the formula is true for \(k + 1\), assuming it is true for \(k\), we have successfully used mathematical induction to prove the given equation: \(1(1!)+2(2!)+\cdots+n(n!)=(n+1)!-1\), for all natural numbers \(n\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Factorial
A factorial, denoted as \(!\), is a concept in mathematics where you multiply a series of descending natural numbers. For any positive integer \(n\), the factorial \(n!\) is defined as the product of all positive integers less than or equal to \(n\). For example:
  • \(1! = 1\)
  • \(2! = 2 \times 1 = 2\)
  • \(3! = 3 \times 2 \times 1 = 6\)
A special case of factorial is \(0!\), which is defined to be \(1\). This definition allows for consistent computation in various mathematical expressions.Factorials are commonly used in permutations, combinations, and other areas of mathematics that require counting arrangements or sequences, such as in the problem we are solving here. In our specific exercise, the factorial is used in expressions like \(n(n!)\), which represents a weighted sum that results in another factorial, \((n+1)!\), demonstrating a beautiful mathematical symmetry.
Base Case
The base case is the starting point of a proof by mathematical induction. It demonstrates that a particular statement holds true for an initial value. In our given exercise, the base case involves verifying the formula for \(n = 1\).Let's break it down:
  • On the left side of the equation, for \(n = 1\), we calculate: \(1(1!) = 1\).
  • On the right side, the expression becomes: \((1+1)! - 1 = 2! - 1 = 2 - 1 = 1\).
Since both sides are equal, the base condition passes. Successful validation of the base case is essential. It creates a foundation for the induction process, acting like the first rung of a ladder which helps in moving upwards to prove the statement for all natural numbers by subsequent steps.
Inductive Hypothesis
The inductive hypothesis is a powerful tool in mathematical induction, where we assume that a given statement or formula holds true for some integer \(k\). This assumption acts as the main bridge to proving the formula for \(k + 1\), thereby extending the truth of the statement to all subsequent integers.In the exercise, our inductive hypothesis is:\[1(1!) + 2(2!) + \cdots + k(k!) = (k+1)! - 1\]This assumption allows us to use what we hope to prove at step \(k+1\) by adding \((k+1)((k+1)!)\) and showing it forms the next part of the series:\[1(1!) + 2(2!) + \cdots + k(k!) + (k+1)((k+1)!) = (k+2)! - 1\]Utilizing the inductive hypothesis helps show that if the formula holds for \(k\), it naturally holds for \((k+1)\). It's much like assuming a bridge can hold a certain load (\(k\)) to reason that it can hold just a bit more (\(k+1\)). This connection between numbers leads to proving the formula valid for all natural numbers.

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