Question

Use the Principle of Mathematical Induction to show that the given statement is true for all natural numbers \(n\). $$ 1^{2}+2^{2}+3^{2}+\cdots+n^{2}=\frac{1}{6} n(n+1)(2 n+1) $$

Short answer

The statement is true for all natural numbers \( n \).

Step-by-step solution

Basis Step

Verify the statement for the initial value, usually \( n = 1 \). Substitute \( n = 1 \) into the left-hand side and right-hand side of the equation and check if both sides are equal. Left-hand side: \( 1^2 = 1 \). Right-hand side: \( \frac{1}{6} \times 1 \times (1+1) \times (2 \times 1 + 1) = \frac{1}{6} \times 1 \times 2 \times 3 = 1 \). Since both sides are equal, the base case is verified.

Inductive Hypothesis

Assume the statement is true for some arbitrary natural number \( k \). That is, assume \( 1^2 + 2^2 + 3^2 + \, \cdots \, + k^2 = \frac{1}{6} k(k+1)(2k+1) \).

Inductive Step

Prove the statement for \( n = k + 1 \) using the inductive hypothesis. Consider \( 1^2 + 2^2 + 3^2 + \, \cdots \, + k^2 + (k+1)^2 \). Using the inductive hypothesis: \( \frac{1}{6} k(k+1)(2k+1) + (k+1)^2 \).

Simplify the Expression

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

Continue Simplifying the Expression

Continue simplifying: \( (k+1) \left[ \frac{1}{6} (2k^2 + k + 6k + 6) \right] = \frac{1}{6} (k+1)(2k^2 + 7k + 6) \). Factor the quadratic: \( = \frac{1}{6} (k+1)(k+2)(2k+3) \). This matches the form of the original statement for \( k+1 \).

Conclusion

Since the statement is true for \( n=1 \) and the inductive step holds, by the Principle of Mathematical Induction, the statement is true for all natural numbers \( n \).

Key concepts

Basis Step

The Principle of Mathematical Induction begins with the basis step, where we verify the given statement for the initial value, most commonly taken as \( n = 1 \).
Let's substitute \( n = 1 \) into the equation's left-hand side:

• Left-hand side: \( 1^2 = 1 \).
• Right-hand side: \( \frac{1}{6} \times 1 \times (1+1) \times (2 \times 1 + 1) = \frac{1}{6} \times 1 \times 2 \times 3 = 1 \).

Since both sides are equal, the base case is verified.
This confirms that the statement is true for \( n = 1 \).
Now, we need to establish that if the statement holds for any natural number \( k \), it also holds for \( k + 1 \).

Inductive Hypothesis

The inductive hypothesis is a critical step in the induction process. Here, we assume that the statement is true for an arbitrary natural number \( k \).
This means we assume:
\[ 1^2 + 2^2 + 3^2 + \cdots + k^2 = \frac{1}{6} k(k+1)(2k+1) \]
This assumption is what we use to prove the next step in the sequence, ensuring the logic holds as we proceed.

Inductive Step

In the inductive step, we must show that if the statement is true for \( n = k \), then it must also be true for \( k + 1 \).
Start by considering the sum up to \( k+1 \):
\[ 1^2 + 2^2 + 3^2 + \cdots + k^2 + (k+1)^2 \]
Using the inductive hypothesis, this can be written as:
\[ \frac{1}{6} k(k+1)(2k+1) + (k+1)^2 \]
You can now factor out \( k+1 \) to simplify further. This factorization and simplification will help us match the form of the original statement, thus proving it for \( k + 1 \).

Simplification of Expressions

The final step involves simplifying the expression \( \frac{1}{6} k(k+1)(2k+1) + (k+1)^2 \):
We factor \( k + 1 \):
\[ \frac{1}{6} (k+1) \left[ k(2k+1) + 6(k+1) \right] \] Continue simplifying inside the brackets: \[ (k+1) \left[ \frac{1}{6} (2k^2 + k + 6k + 6) \right] = \frac{1}{6} (k+1)(2k^2 + 7k + 6) \]
Finally, factorizing the quadratic expression gives us: \[ = \frac{1}{6} (k+1)(k+2)(2k+3) \]
This matches the formula we are aiming to prove.
Therefore, by demonstrating that the statement holds for \( k + 1 \), we complete our proof using the Principle of Mathematical Induction.