How do you find the sum given #Sigma(2i+1)# from i=1 to 5? Calculus Introduction to Integration Sigma Notation 1 Answer Steve M Oct 31, 2016 # sum_(i=1)^(i=5)(2i+1) = 35 # Explanation: # sum_(i=1)^(i=5)(2i+1) = (2(1)+1) + (2(2)+1) + (2(3)+1) + (2(4)+1) + (2(5)+1) # # :. sum_(i=1)^(i=5)(2i+1) = (2+1) + (4+1) + (6+1) + (8+1) + (10+1) # # :. sum_(i=1)^(i=5)(2i+1) = 3 + 5 + 7 + 9 + 11 # # :. sum_(i=1)^(i=5)(2i+1) = 35 # Answer link Related questions How does sigma notation work? How do you use sigma notation to represent the series #1/2+1/4+1/8+…#? Use summation notation to express the sum? What is sigma notation for an arithmetic series with first term #a# and common difference #d# ? How do you evaluate the sum represented by #sum_(n=1)^5n/(2n+1)# ? How do you evaluate the sum represented by #sum_(n=1)^(8)1/(n+1)# ? How do you evaluate the sum represented by #sum_(n=1)^(10)n^2# ? What is sigma notation for a geometric series with first term #a# and common ratio #r# ? What is the value of #1/n sum_{k=1}^n e^{k/n}# ? Question #07873 See all questions in Sigma Notation Impact of this question 9737 views around the world You can reuse this answer Creative Commons License