Question

Use the Principle of Mathematical Induction to show that the given statement is true for all natural numbers \(n\). $$ 3+4+5+\cdots+(n+2)=\frac{1}{2} n(n+5) $$

Short answer

The statement is true for all by induction.

Step-by-step solution

Base Case

Verify the statement for the smallest natural number, which is usually . For this statement, check the case when = 1.

Statement for n = 1

Substitute = 1 in the left-hand side of the equation: 3. Verify the right-hand side: \[\frac{1}{2} \cdot 1 \cdot (1 + 5) = \frac{1}{2} \cdot 1 \cdot 6 = 3.\] Therefore, the statement is true for = 1.

Inductive Hypothesis

Assume the statement is true for some natural number = k. That is, assume \[3 + 4 + 5 + \cdots + (k+2) = \frac{1}{2} k (k+5).\]

Statement for n = k + 1

Prove the statement for = k + 1. Start with the sum up to = k and add the next term: \[3 + 4 + 5 + \cdots + (k+2) + (k+3).\]

Substitute the Induction Hypothesis

Using the inductive hypothesis in step 3, substitute: \[\frac{1}{2} k (k+5) + (k+3).\]

Simplify the Expression

Combine the terms: \[\frac{1}{2} k (k+5) + 2 (k+3) = \frac{1}{2} k (k+5) + \frac{2(k+3)}{1} = \frac{1}{2} k (k+5) + \frac{4(k+3)}{2}.\] Factor out 1/2: \[\frac{1}{2} [k (k+5) + 4 (k+3)].\]

Expand and Combine Like Terms

Expand and combine the terms inside the brackets: \[\frac{1}{2} (k^2 + 5k + 4k + 12) = \frac{1}{2} (k^2 + 9k + 12).\]

Verify Final Expression

Recognize that \[\frac{1}{2} (k^2 + 9k + 12) = \frac{1}{2} (k+1)(k+6).\] Therefore, \[3 + 4 + 5 + \cdots + (k+3) = \frac{1}{2} (k+1) ((k+1)+5).\]

Conclusion

Since the base case and the inductive step are both verified, by the Principle of Mathematical Induction, the statement is true for all natural numbers .

Key concepts

Base Case

To prove a statement by mathematical induction, we first need to verify the statement for the smallest natural number, usually 1. This step is known as the base case. In our given problem, we start by checking when n=1:

For the left-hand side (LHS):
3
For the right-hand side (RHS):
\[\frac{1}{2} \times 1 \times (1+5) = \frac{1}{2} \times 1 \times 6 = 3\]
Both the LHS and the RHS are equal to 3, verifying that the statement holds true for n=1. This is an essential first step as it confirms the starting point of our induction.

Inductive Hypothesis

In the inductive hypothesis, we assume that the statement is true for some arbitrary natural number k. This Assumption is like taking a step on a ladder to reach the next step. Let's assume that the given equation holds true for n = k:

\[3 + 4 + 5 + \cdots + (k+2) = \frac{1}{2} k (k+5)\]

This assumption is the foundation for proving that the statement holds true for the next natural number, n=k+1.

Inductive Step

In the inductive step, we prove that if the statement is true for n=k, then it is also true for n=k+1. We begin by adding the next term in the sequence to our inductive hypothesis:
\[3 + 4 + 5 + \cdots + (k+2) + (k+3)\].
By substituting the inductive hypothesis, this becomes:
\[\frac{1}{2} k (k+5) + (k+3)\].
Next, we simplify the expression:
\[\frac{1}{2} k (k+5) + 2 (k+3) = \frac{1}{2} k (k+5) + \frac{4 (k+3)}{2} = \frac{1}{2}[k (k+5) + 4 (k+3)\].
Expanding and combining like terms inside the brackets, we get:
\[\frac{1}{2} (k^2 + 5k + 4k + 12) = \frac{1}{2} (k^2 + 9k + 12)\].
Finally, recognizing the simplified form, we see:
\[3 + 4 + 5 + \cdots + (k+3) = \frac{1}{2} (k+1) ((k+1) + 5)\].
Therefore, the statement holds true for n=k+1 if it is true for n=k.

Summation Formula

A summation formula provides a concise way to express the sum of a sequence. In this exercise, the sequence 3 + 4 + 5 + ... + (n+2) is given in its general form. We are to show it equals:

\[\frac{1}{2} n (n+5)\].

Summation formulas are crucial in problems requiring simplification of sequences. They offer more efficient and understandable solutions compared to adding terms individually. The principle of mathematical induction is an ideal method to prove the validity of these formulas for all natural numbers. Using the summation formula, we established that each sequence's sum holds true from the base case to an inductive step, thereby covering all possible natural numbers.