Math Problem Statement

Solve by mathematical induction: P(n): 1 + 2 + 3 + ... + n + (n + 1) + (n + 2) = (n + 2)(n + 3) / 2

Solution

The problem you've provided is asking to prove the given statement using mathematical induction. Here's the breakdown of the problem:

Statement to Prove (P(n)):

1+2+3++n+(n+1)+(n+2)=(n+2)(n+3)21 + 2 + 3 + \ldots + n + (n + 1) + (n + 2) = \frac{(n+2)(n+3)}{2} This expression represents the sum of the first n+2n+2 integers, and we need to prove that it equals the formula on the right-hand side using mathematical induction.

Steps of Mathematical Induction:

  1. Base Case: Verify the statement for a specific small value of nn, usually n=1n = 1.
  2. Inductive Hypothesis: Assume the statement is true for some n=kn = k.
  3. Inductive Step: Prove that if the statement holds for n=kn = k, it also holds for n=k+1n = k + 1.

Let's go step-by-step:


Step 1: Base Case ( n=1n = 1 )

For n=1n = 1, the left-hand side becomes: 1+2+3=61 + 2 + 3 = 6 Now, the right-hand side is: (1+2)(1+3)2=3×42=6\frac{(1+2)(1+3)}{2} = \frac{3 \times 4}{2} = 6 So, the base case holds for n=1n = 1.


Step 2: Inductive Hypothesis

Assume the statement is true for n=kn = k. This means: 1+2+3++k+(k+1)+(k+2)=(k+2)(k+3)21 + 2 + 3 + \ldots + k + (k+1) + (k+2) = \frac{(k+2)(k+3)}{2}


Step 3: Inductive Step

We now need to prove that the statement holds for n=k+1n = k + 1. That is, we need to show: 1+2+3++(k+1)+(k+2)+(k+3)=(k+3)(k+4)21 + 2 + 3 + \ldots + (k+1) + (k+2) + (k+3) = \frac{(k+3)(k+4)}{2} Start by adding (k+3)(k + 3) to both sides of the inductive hypothesis: (1+2+3++k+(k+1)+(k+2))+(k+3)=(k+2)(k+3)2+(k+3)\left( 1 + 2 + 3 + \ldots + k + (k+1) + (k+2) \right) + (k+3) = \frac{(k+2)(k+3)}{2} + (k+3) Factor the right-hand side: =(k+3)((k+2)2+1)= (k+3) \left( \frac{(k+2)}{2} + 1 \right) =(k+3)(k+2+22)= (k+3) \left( \frac{k+2 + 2}{2} \right) =(k+3)(k+42)= (k+3) \left( \frac{k+4}{2} \right) =(k+3)(k+4)2= \frac{(k+3)(k+4)}{2} Thus, the statement holds for n=k+1n = k + 1.


Conclusion:

By mathematical induction, the statement is true for all n1n \geq 1.


Would you like further details on any step or have any questions?

Here are 5 related questions to explore further:

  1. How does the formula for the sum of the first nn integers relate to the inductive proof?
  2. Can mathematical induction be used to prove other types of sequences, such as geometric ones?
  3. How can you apply mathematical induction to inequalities?
  4. What happens if the base case fails in mathematical induction?
  5. How do we identify the right base case for different types of problems?

Tip: Always double-check your inductive hypothesis and base case. These are crucial for a correct proof!

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Math Problem Analysis

Mathematical Concepts

Mathematical Induction
Sum of Integers

Formulas

Sum of first k integers: 1 + 2 + ... + k = k(k+1)/2
Formula to prove: 1 + 2 + ... + n + (n + 1) + (n + 2) = (n+2)(n+3)/2

Theorems

Principle of Mathematical Induction

Suitable Grade Level

High School (Grades 10-12), Early College