Using principle of mathematical induction prove that 1^2 + 2^2 + 3^2 + ..... + n^2 = n/6(n + 1)(2n + 1)

To prove that:

\[
1^2 + 2^2 + 3^2 + \dots + n^2 = \frac{n}{6}(n+1)(2n+1)
\]

we will use the **Principle of Mathematical Induction (PMI)**.

### Steps in PMI:

1. **Base Case:** Verify the formula is true for \( n = 1 \).
2. **Inductive Hypothesis:** Assume the formula is true for \( n = k \), i.e.,
   \[
   1^2 + 2^2 + 3^2 + \dots + k^2 = \frac{k}{6}(k+1)(2k+1).
   \]
3. **Inductive Step:** Prove the formula is true for \( n = k+1 \), i.e.,
   \[
   1^2 + 2^2 + 3^2 + \dots + k^2 + (k+1)^2 = \frac{k+1}{6}(k+2)(2k+3).
   \]

---

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

For \( n = 1 \), the left-hand side is:
\[
1^2 = 1.
\]

The right-hand side is:
\[
\frac{1}{6}(1+1)(2 \cdot 1 + 1) = \frac{1}{6}(2)(3) = 1.
\]

Thus, the formula holds for \( n = 1 \).

---

### Step 2: Inductive Hypothesis:

Assume the formula is true for \( n = k \):
\[
1^2 + 2^2 + 3^2 + \dots + k^2 = \frac{k}{6}(k+1)(2k+1).
\]

---

### Step 3: Inductive Step:

We must show that:
\[
1^2 + 2^2 + 3^2 + \dots + k^2 + (k+1)^2 = \frac{k+1}{6}(k+2)(2k+3).
\]

Using the inductive hypothesis, the left-hand side becomes:
\[
\left( \frac{k}{6}(k+1)(2k+1) \right) + (k+1)^2.
\]

Factor \( (k+1) \) from both terms:
\[
= \frac{k}{6}(k+1)(2k+1) + (k+1)^2.
\]

\[
= (k+1) \left( \frac{k}{6}(2k+1) + (k+1) \right).
\]

Simplify the terms inside the parentheses:
\[
\frac{k}{6}(2k+1) + (k+1) = \frac{1}{6}\big[k(2k+1) + 6(k+1)\big].
\]

Expand:
\[
= \frac{1}{6}\big[2k^2 + k + 6k + 6\big].
\]

\[
= \frac{1}{6}(2k^2 + 7k + 6).
\]

Factorize \( 2k^2 + 7k + 6 \):
\[
= \frac{1}{6}(k+2)(2k+3).
\]

Substitute back:
\[
= (k+1) \cdot \frac{1}{6}(k+2)(2k+3).
\]

\[
= \frac{k+1}{6}(k+2)(2k+3).
\]

---

### Conclusion:

By the principle of mathematical induction, the formula:
\[
1^2 + 2^2 + 3^2 + \dots + n^2 = \frac{n}{6}(n+1)(2n+1)
\]
is true for all \( n \geq 1 \).

---

Would you like further clarifications or more examples?

---

### 5 Related Questions:
1. How do you prove the sum of cubes formula using mathematical induction?
2. Can you derive the formula for the sum of the first \( n \) cubes geometrically?
3. How does mathematical induction work for inequalities, such as proving \( 2^n > n^2 \) for \( n \geq 5 \)?
4. What are some common pitfalls to avoid when using mathematical induction?
5. How can you use induction to prove properties of divisibility?

### Tip:
When working with induction, always double-check the base case—it’s the foundation of the entire proof!

Learn how to prove the formula for the sum of squares of the first n integers using the principle of mathematical induction. This step-by-step solution includes all necessary steps, from the base case to the inductive step, suitable for high school students (Grades 9-12).

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation