ing your knowledge of series: Prove that the sum of the first \( n \), odd natural numbers is \( n^{2} \)

To prove that the sum of the first \( n \) odd natural numbers is \( n^{2} \), we can use **mathematical induction**. Here's a step-by-step proof: ### **Statement to Prove** \[ \sum_{k=1}^{n} (2k - 1) = n^{2} \] In other words: \[ 1 + 3 + 5 + \dots + (2n - 1) = n^{2} \] ### **Base Case** First, verify the statement for \( n = 1 \): \[ \text{Left Side (LS)} = 1 \] \[ \text{Right Side (RS)} = 1^{2} = 1 \] Since LS = RS, the base case holds. ### **Inductive Step** Assume the statement holds for some arbitrary positive integer \( k \); that is: \[ \sum_{i=1}^{k} (2i - 1) = k^{2} \] **Goal:** Show that the statement holds for \( k + 1 \), i.e., \[ \sum_{i=1}^{k+1} (2i - 1) = (k + 1)^{2} \] **Starting from the induction hypothesis:** \[ \sum_{i=1}^{k+1} (2i - 1) = \left( \sum_{i=1}^{k} (2i - 1) \right) + (2(k+1) - 1) \] \[ = k^{2} + (2k + 2 - 1) \quad \text{(by the induction hypothesis)} \] \[ = k^{2} + (2k + 1) \] \[ = k^{2} + 2k + 1 \] \[ = (k + 1)^{2} \] This shows that if the statement holds for \( k \), it also holds for \( k + 1 \). ### **Conclusion** By the principle of mathematical induction, the sum of the first \( n \) odd natural numbers is indeed \( n^{2} \) for all positive integers \( n \). ### **Alternative Proof: Using Arithmetic Series Formula** Alternatively, we can use the formula for the sum of an arithmetic series. The first \( n \) odd natural numbers form an arithmetic sequence with: - First term (\( a_1 \)) = 1 - Common difference (\( d \)) = 2 - Number of terms (\( n \)) = \( n \) The \( n \)-th term (\( a_n \)) is: \[ a_n = a_1 + (n - 1)d = 1 + (n - 1) \times 2 = 2n - 1 \] The sum of an arithmetic series is given by: \[ S_n = \frac{n}{2} (a_1 + a_n) = \frac{n}{2} (1 + (2n - 1)) = \frac{n}{2} \times 2n = n^{2} \] This confirms that the sum of the first \( n \) odd natural numbers is \( n^{2} \). ### **Visual Proof** There's also a beautiful visual proof where arranging dots in squares demonstrates that each new layer adds an odd number of dots, resulting in a complete square of size \( n \times n \).  In this diagram, each successive layer adds an odd number of dots (1, 3, 5, ...) to form a larger square. ### **Summary** Whether through mathematical induction, the arithmetic series formula, or a visual arrangement, we have proven that: \[ 1 + 3 + 5 + \dots + (2n - 1) = n^{2} \]