P(n):1+3+5+...+2n−1=n2The statement P(n) is
P(n):1+3+5+...+2n−1=n2P(1):1=12⇒1=1− trueAssume P(k) is true1+3+5+...+2k−1=k2Now, P(k+1):1+3+5+...+2n−1+2k+1=k2+2k+1=(k+1)2P(k+1) is also true.Hence, P(n) is true for all natural numbers
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