證明

N o . 1 : \mathcal{No.}1: No.1: 證: 設 n = 2 ∗ k + 1 n=2*k+1 n=2∗k+1 ∴ n 2 − 1 = ( 2 ∗ k + 1 ) 2 − 1 = 4 ∗ k 2 + 4 ∗ k = 4 ∗ ( k ∗ ( k + 1 ) ) \therefore n^2-1\\=(2*k+1)^2-1\\=4*k^2+4*k\\=4*(k*(k+1)) ∴n

>>阅读原文<<