Arrange the given steps in the correct order to prove that 1-2 1 2-3 + ...+ = k- (k+1 ) is true for all n > 0 using the concept of mathematical induction. Rank the options below. By adding (kunj(k 2) (kijk ?) on both the sides, we get 12 12 + 23 23 + ... + K . ( k + 1 ) K . ( k + 1 ) ( k + 1 ) ( k + 2 ) ( k + 1 ) ( k + 2 ) - K+I k+ I ( k+ 1) (k+2) ( k+ 1) (k+2) We have completed both the basis step and the inductive step; so, by the principle of mathematical induction, the statement is true for every positive integer n. Assume that for some k > 0, 1 k k 5 + 23 23 k. (k+ 1) k (k+1) k(k+ 2) +1 Therefore, we have To + 23 + ... + K( k+1 ) + ( *+1) (1 +2) (k+ 1)(k+ 2) 1 + ... + k(k+ 2) + 1 1.2 k-(k+1 ) + ( ke + 1) (k +2) (k+ 1) (k+ 2) This is true because each term on left hand-side = ke -, by the inductive hypothesis. Therefore, k(k42)+1 k242k+1 kill k(k+2)+1 k242k+1 k+1 ( k+1) (k+2) (k+1) (142) k42 (k+1) (k+2) (k +1) (k+2) For n = 1, the left-hand side of the theorem is 12 12 = - and the right-hand side = ntl n+1 -= 1