Write the following series in sigma notation: 3 4 + 4 5 + 5 6

Given, 3 4 + 4 5 + 5 6 This is a finite arithmetic series which can be rewritten as: 3 3 + 1 + 4 4 + 1 + 5 5 + 1 Thus, the general term of this series is n n + 1 , where n = 3 , 4 , and 5 . Since the first term is obtained by substituting n = 3 in the general term n n + 1 . And, 5 is the last value of n that gives us the last term. So, in sigma notation we take the lower limit as 3 , while upper limit as 5 . Therefore, the sigma notation of the given series is as follows: 3 4 + 4 5 + 5 6 = ∑ n = 3 5 n n + 1

Write the following series in sigma notation: - 1 + 1 2 - 1 3 + 1 4 - ⋯ + 1 100

Given, - 1 + 1 2 - 1 3 + 1 4 - ⋯ + 1 100 This is a finite arithmetic series which can be rewritten as: ( - 1 ) 1 1 1 + ( - 1 ) 2 1 2 + ( - 1 ) 3 1 3 + ( - 1 ) 4 1 4 + … + ( - 1 ) 100 1 100 Observe that the denominator of each term in the above series is a natural number from 1 to 100 , and its numerator is 1 . Also, observe that the first term is negative, and the signs of following terms alternate. Therefore, the general term of the given series can be taken as - 1 n 1 n , where n = 1 , 2 , 3 , . . . . , 100 . Since the first term is obtained by substituting n = 1 in the general term - 1 n 1 n . And, 100 is the last value of n that gives us the last term. So, in sigma notation we take the lower limit as 1 , while upper limit as 100 . Therefore, the sigma notation of the given series is as follows: - 1 + 1 2 - 1 3 + 1 4 - ⋯ + 1 100 = ∑ n = 1 100 - 1 n 1 n