Math Problem Statement

Using the principle of mathematical induction, prove the following statements for all positive integers: (a) 1 + 2 + 3 + ... + n = n(n + 1) / 2 (b) 1·2 + 2·3 + 3·4 + ... + n(n + 1) = n(n + 1)(n + 2) / 3

Solution

Let's solve this by proving each part using the principle of mathematical induction.

Part (a)

Statement:

For all positive integers nn: 1+2+3++n=n(n+1)21 + 2 + 3 + \ldots + n = \frac{n(n+1)}{2}

Proof by Induction:

  1. Base Case (n=1n = 1): 1=1(1+1)2=122=11 = \frac{1 \cdot (1 + 1)}{2} = \frac{1 \cdot 2}{2} = 1 The base case holds true.

  2. Inductive Step:

    • Assume the formula holds for some integer kk, i.e., 1+2+3++k=k(k+1)21 + 2 + 3 + \ldots + k = \frac{k(k+1)}{2}
    • Now, prove that it holds for k+1k + 1, i.e., show that: 1+2+3++k+(k+1)=(k+1)(k+2)21 + 2 + 3 + \ldots + k + (k + 1) = \frac{(k + 1)(k + 2)}{2}
    • Starting with the inductive hypothesis: 1+2+3++k+(k+1)=k(k+1)2+(k+1)1 + 2 + 3 + \ldots + k + (k + 1) = \frac{k(k+1)}{2} + (k + 1)
    • Factor out (k+1)(k + 1) from the right side: =k(k+1)+2(k+1)2=(k+1)(k+2)2= \frac{k(k+1) + 2(k+1)}{2} = \frac{(k+1)(k+2)}{2}
    • This matches the required expression for k+1k + 1.

Since both the base case and inductive step are proven, the statement holds for all positive integers nn.


Part (b)

Statement:

For all positive integers nn: 12+23+34++n(n+1)=n(n+1)(n+2)31 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \ldots + n(n+1) = \frac{n(n+1)(n+2)}{3}

Proof by Induction:

  1. Base Case (n=1n = 1): 12=1233=63=21 \cdot 2 = \frac{1 \cdot 2 \cdot 3}{3} = \frac{6}{3} = 2 The base case holds true.

  2. Inductive Step:

    • Assume the formula holds for some integer kk, i.e., 12+23+34++k(k+1)=k(k+1)(k+2)31 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \ldots + k(k+1) = \frac{k(k+1)(k+2)}{3}
    • Now, prove that it holds for k+1k + 1, i.e., show that: 12+23+34++k(k+1)+(k+1)(k+2)=(k+1)(k+2)(k+3)31 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \ldots + k(k+1) + (k + 1)(k + 2) = \frac{(k + 1)(k + 2)(k + 3)}{3}
    • Starting with the inductive hypothesis: 12+23+34++k(k+1)+(k+1)(k+2)=k(k+1)(k+2)3+(k+1)(k+2)1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \ldots + k(k+1) + (k + 1)(k + 2) = \frac{k(k+1)(k+2)}{3} + (k + 1)(k + 2)
    • Factor out (k+1)(k+2)(k + 1)(k + 2) from the right side: =k(k+1)(k+2)+3(k+1)(k+2)3=(k+1)(k+2)(k+3)3= \frac{k(k+1)(k+2) + 3(k+1)(k+2)}{3} = \frac{(k+1)(k+2)(k+3)}{3}
    • This matches the required expression for k+1k + 1.

Since both the base case and inductive step are proven, the statement holds for all positive integers nn.


Would you like further details or explanations?

Here are some related questions for practice:

  1. What is the principle of mathematical induction?
  2. How would you prove a different series sum using induction?
  3. Can induction be used to prove inequalities?
  4. What are common errors to avoid in inductive proofs?
  5. How does induction differ from other proof techniques?

Tip: When using induction, always start with a clear base case and ensure each step logically follows from the assumption.

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Math Problem Analysis

Mathematical Concepts

Mathematical Induction
Summation of Series

Formulas

Sum of first n natural numbers: S = n(n + 1) / 2
Sum of first n terms of sequence of products: S = n(n + 1)(n + 2) / 3

Theorems

Principle of Mathematical Induction

Suitable Grade Level

Grades 11-12