mathematical induction

prove the following statements by mathematical induction,where n is a natural number.

(n+1)2+(n+2)2+(n+3)2+...+(2n)2 = [n(2n+1)(7n+1)] / 6
當 n = 1
LHS = (1 + 1)2 = 4
RHS = [n(2n+1)(7n+1)] / 6
= [1(2(1)+1)(7(1)+1)] / 6
= [1(3)(8)] / 6
= 4
所以n=1時成立
設 n = k 時成立，
則
(k+1)2+(k+2)2+(k+3)2+...+(2k)2 = [k(2k+1)(7k+1)] / 6
當n=k + 1
LHS = ((k+1)+1)2+((k+1)+2)2+((k+1)+3)2+...+(2(k+1))2
= (k+2)2 + (k+3)2 + (k+4)2 + …. (2k)2 + (2k+1)2 + (2k+2)2
= (k+1)2 + (k+2)2 + (k+3)2 + (k+4)2 + …. (2k)2 + (2k+1)2 + (2k+2)2 – (k+1)2
= [k(2k+1)(7k+1)] / 6 + (2k+1)2 + (2k+2)2 – (k+1)2
= [k(2k+1)(7k+1)] / 6 + 6(2k+1)2/6 + 3(k+1)2
= (2k+1)[k(7k+1) + 6(2k+1)]/6 + 3(k+1)2
= (2k+1)[7k2+k + 12k+6)]/6 + 3(k+1)2
= (2k+1)[7k2+13k+6)]/6 + 3(k+1)2
= (2k+1)(7k+6)(k+1)/6 + 18(k+1)2/6
= (2k+1)(7k+6)(k+1)/6 + 18(k+1)2/6
= (k+1)[(2k+1)(7k+6) + 18(k+1)]/6
= (k+1)[14k2+19k+6 + 18k+18)]/6
= (k+1)[14k2+37k+24]/6
= (k+1)(2k+3)(7k+8)/6
= (k+1)[2(k+1)+1][7(k+1)+1]/6
=RHS
所以當 n 為任何值時這式都成立。