Question

Several algorithms have a loop structure that iterates n times and for the k-th iteration, they perform on the order of k2 operations. In this case, the running time of the loop can be obtained by k2. In this problem, we will derive a closed formula for this sum and show that the sum is O(ui) 1. A telescoping sum is a sum of the form Σ-i (aj - a-1). It is easy to see that this sum equals an -do (try it for example for j 4). In class, we stated that ΣΑι k-n(n + 1 )(2n + 1 )/6. Derive this formula using the telescoping sum. Hint: Take ak-: k 3 in the telescoping sum. 2. Show that () by explicitly finding the values of the constants k and C and demonstrating the result.

Answer 1

1) Telescopic stu n: 1) (2n ) 6 k-1 Proof: The above can be proved by using the principle of mathematical induction. For example, N-1:1211*(1+1)* (2*1+1)/6. It is true for the values such that 12+ 242+312+N2 N*N1) (2N+1)/6, Add (N+12 on both sides (12+ 242+312+ +N2)+ (N+12 N*N+1) (2N+1)/6 +(N+1)2, 1^2+ 2"2 + 3"2 + + N^2+ (N-1)^2 = (N+1)* [N*(2*N-1)/G+ N +1], N+1) 2*N 2+7*N+6)/6, (N+1)*[(N-IHI]*[2*(N-1)+1]/6. This is the statement need to prove for "N+1". Apply here the principle of mathematical induction to conclude the statement for all the values N>1.

The another approach is that a(n)= N*(N+1)*(2N+1)/6. N*L(2 N 2+3*N+1)-(2 NA2-3*N+1)]/6, N* (69)/6. Let apply, 112 f(1) - f(0), 22 f(2) - f(1), 3n2 f(3) f(2), (N-1)^2-a(N-1)-a(N-2), While adding all those together, there will be cancellation of terms on right side. After cancellation, the term will be 12 22 +32 +... + N2-a(n)-a(0)- N*(N+1) (2N+1)/6 This is a big cancellation that occurs in a sum and this type of cancellation is referred as "telescoping sum". From the above steps, the telescopic sum of first N squares 1) (2n ) 6

2) It is need to prove that k 1 Since, result of telescopic sum of first N squares results in Ση_1 k-n (n + 1) (2n +1) / 6 While multiplying the terms on right side in above result of telescopic sum, it will results in n2 N*N-) 2N+1)6" polynomial which is higher order term in the polynomial expression This higher order notation is to represent the time complexity. So, according to big-oh notation, the running time for telescopic sum of first "N" squares is O(n3) Hence, it is proved O(n') k-1