Prove by induction on n that, for all positive integers n: n (i(i!) = (n+1)! –1 %3D i=1

Advanced Engineering Mathematics
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ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
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**Inductive Proof Exercise**

**Problem Statement:**
Prove by induction on \( n \) that, for all positive integers \( n \):

\[
\sum_{i=1}^{n} \left( i(i!) \right) = (n+1)! - 1
\]

**Explanation:**

This exercise requires the use of mathematical induction to prove the given equation involving summation and factorials for all positive integers \( n \).

- **Summation Notation:** 
  \(\sum_{i=1}^{n} \left( i(i!) \right)\) represents the sum of the expression \( i(i!) \) as \( i \) ranges from 1 to \( n \). Here, \( i! \) denotes the factorial of \( i \), which is the product of all positive integers less than or equal to \( i \).

- **Factorial Notation:** 
  \((n+1)! \) denotes the factorial of \( n+1 \), which is the product of all positive integers less than or equal to \( n+1 \).

- **Proof by Induction:**
  1. **Base Case:** Verify the statement for the initial value \( n = 1 \).
  2. **Inductive Step:** Assume the statement is true for \( n = k \) (Inductive Hypothesis) and then prove it for \( n = k + 1 \).

By completing both steps, you can conclude that the statement holds for all positive integers \( n \).
Transcribed Image Text:**Inductive Proof Exercise** **Problem Statement:** Prove by induction on \( n \) that, for all positive integers \( n \): \[ \sum_{i=1}^{n} \left( i(i!) \right) = (n+1)! - 1 \] **Explanation:** This exercise requires the use of mathematical induction to prove the given equation involving summation and factorials for all positive integers \( n \). - **Summation Notation:** \(\sum_{i=1}^{n} \left( i(i!) \right)\) represents the sum of the expression \( i(i!) \) as \( i \) ranges from 1 to \( n \). Here, \( i! \) denotes the factorial of \( i \), which is the product of all positive integers less than or equal to \( i \). - **Factorial Notation:** \((n+1)! \) denotes the factorial of \( n+1 \), which is the product of all positive integers less than or equal to \( n+1 \). - **Proof by Induction:** 1. **Base Case:** Verify the statement for the initial value \( n = 1 \). 2. **Inductive Step:** Assume the statement is true for \( n = k \) (Inductive Hypothesis) and then prove it for \( n = k + 1 \). By completing both steps, you can conclude that the statement holds for all positive integers \( n \).
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