The question asks to prove that:

$1^2 + 3^2 + 5^2 + \cdots + (2n+1)^2 = \frac{(n+1)(2n+1)(2n+3)}{3}$ for any nonnegative integer $n$.

Step-by-Step Proof

1. Sum of Odd Squares: The expression on the left-hand side is the sum of the squares of the first $n+1$ odd numbers. Odd numbers can be written as $2k + 1$ for integers $k$. So the sequence we are summing is: $(2(0) + 1)^2, (2(1) + 1)^2, (2(2) + 1)^2, \dots, (2n+1)^2$ This is equivalent to: $1^2 + 3^2 + 5^2 + \dots + (2n + 1)^2$

2. Recognizing a Formula: We need to derive or recognize the formula for the sum of squares of odd integers. From known summation formulas, the sum of the squares of the first $n+1$ odd numbers is: $\sum_{k=0}^{n} (2k+1)^2$ Expanding each term in the sum: $(2k+1)^2 = 4k^2 + 4k + 1$ So the sum becomes: $\sum_{k=0}^{n} (4k^2 + 4k + 1) = 4 \sum_{k=0}^{n} k^2 + 4 \sum_{k=0}^{n} k + \sum_{k=0}^{n} 1$

3. Using Known Summation Formulas: We can now use standard summation formulas to simplify each part:

   • $\sum_{k=0}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$
   • $\sum_{k=0}^{n} k = \frac{n(n+1)}{2}$
   • $\sum_{k=0}^{n} 1 = n+1$

4. Putting It All Together: Substituting these into the expanded sum: $4 \left(\frac{n(n+1)(2n+1)}{6}\right) + 4 \left(\frac{n(n+1)}{2}\right) + (n+1)$ Simplify each term: $\frac{2n(n+1)(2n+1)}{3} + 2n(n+1) + (n+1)$ Factor $(n+1)$ from all terms: $(n+1) \left( \frac{2n(2n+1)}{3} + 2n + 1 \right)$ Combine terms in the parentheses: $(n+1) \left( \frac{2n(2n+1) + 6n + 3}{3} \right) = (n+1) \left( \frac{4n^2 + 8n + 3}{3} \right)$ Finally, simplify to get: $\frac{(n+1)(2n+1)(2n+3)}{3}$

This proves the formula for the sum of the squares of the first $n+1$ odd numbers.

Would you like a deeper explanation on any part of this proof?

Related Questions:

1. Can you explain how to derive the sum formulas for $\sum k^2$ and $\sum k$?
2. How does factoring out $(n+1)$ help in simplifying the expression?
3. What other summation formulas are commonly used in proving identities like this?
4. How does the sum of squares of odd numbers differ from the sum of squares of even numbers?
5. Could this proof be done using mathematical induction?

Tip: Always try to simplify terms and factor common elements when handling summations; it makes managing large expressions much easier.