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., $1^2 + 2^2 + 3^2 + \dots + k^2 + (k+1)^2 = \frac{k+1}{6}(k+2)(2k+3).$

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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 $n = 1$.

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Step 2: Inductive Hypothesis:

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

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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: $= (k+1) \cdot \frac{1}{6}(k+2)(2k+3).$

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

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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 $n \geq 1$.

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Would you like further clarifications or more examples?

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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!