Solve using mathematical induction Share with your friends Share 0 Lovina Kansal answered this Dear student 12+32+52+...+(2n-1)2=n3(2n-1)(2n+1)⇒12+32+52+...+(2n-1)2=n3(4n2-1)Let P(n) be the given statementNow,P(n)=12+32+52+...+(2n-1)2=n3(4n2-1)STEP 1: P(1)=12=1=13×412-1Hence, P(1) is true.STEP 2: Let P(m) be true. 12+32+52+...+(2m-1)2=m3(4m2-1)To prove: P(m+1) is true whenever P(m) is true.i.e.,12+32+52+...+(2m+1)2=m+134(m+1)2-1We know that P(m) is true.Thus, we have12+32+52+...+(2m-1)2=m3(4m2-1)⇒12+32+52+...+(2m-1)2+(2m+1)2=m3(4m2-1)+(2m+1)2⇒P(m+1)=134m3-m+12m2+3+12m⇒P(m+1)=134m3-m+8m2+4m+4m2+8m+3⇒P(m+1)=134m3+8m2+3m+4m2+8m+3⇒P(m+1)=13m(4m2+8m+3)+1(4m2+8m+3)⇒P(m+1)=13(m+1)(4m2+8m+3)⇒P(m+1)=13(m+1)(4m2+2m+1-1)⇒P(m+1)=13m+14m+12-1Thus, P(m+1) is trueBy PMI,P(n) is true for all n∈N Regards 1 View Full Answer