Word Count : 565

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We are tasked with proving the following statement by induction: \[ 1^2 + 3^2 + 5^2 + \dots + (2n-1)^2 = \frac{1}{3} n (4n^2 - 1), \quad \forall n \in \mathbb{N}. \] ### Step 1: Base Case (n = 1) For \( n = 1 \), the left-hand side (LHS) is: \[ 1^2 = 1. \] The right-hand side (RHS) is: \[ \frac{1}{3} \times 1 \times (4 \times 1^2 - 1) = \frac{1}{3} \times 1 \times (4 - 1) = \frac{1}{3} \times 3 = 1. \] Since LHS = RHS, the base case holds. ______ ___ ________ _______ _______.
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