Data Structures, Algorithms, & Applications in C++
Chapter 3, Exercise 15

(E5)
(sum from 1 to n)ik <= (sum from 1 to n)nk = nk+1. Therefore, (sum from 1 to n)ik = O(nk+1). Further, (sum from 1 to n)ik >= (sum from n/2 to n)(n/2)k = (n/2)k+1. Therefore, (sum from 1 to n)ik = Omega(nk+1). (Note that 2k+1 is a constant as k is a constant.) Consequently, (sum from 1 to n)ik = Theta(nk+1).

(E6)
(sum from 1 to n)ri = (rn+1-1)/(r-1) - 1 = Theta(rn). From this it follows that (sum from 1 to n)ri = O(rn) and (sum from 1 to n)ri = Omega(rn).