The correct option is C P(n) ⇒P(n+1)
Consider P(1)LHS=1 and RHS=3+1=4So, LHS≠RHS for n=1Given P(n): 1+3+5+...+(2n−1)=3+n2Assume P(k) is true,i.e., 1+3+5+...+(2k−1)=3+k2 ...(1)Consider 1+3+5+...+(2k−1)+(2(k+1)−1) =1+3+5+...+(2k−1)2+(2k+1) =3+k2+(2k+1) ...(Using (1)] =3+(k+1)2Thus, P(k+1) is true when P(k) is true.Although P(1) is not true, assuming P(k) tobe true implies P(k+1) is true.