Question

i) Write a Sigma-notation summation for: the sum of the first “n” odd positive integers. For example, if n=4, it should sum like this: 1 + 3 + 5 + 7.

ii) Starting with: 1+sum r^n, n=1 to infinity, rewrite it as just one summation without the 1+ out front.

iii) Starting with: sum 1/n, n=1 to infinity, rewrite it as two terms out front, and then the sum starting at n=3.

iv) Starting with: sum 1/(n+1), n=0 to 5, rewrite it to start with n=1. Try thinking of it as a u-substitution? Here, we aren’t talking about moving terms out front--keep it as one single sum.

Answer 1

i nth odd positive integer is (2n-1). Therefore sum of first a positive odd integers is Ž (22-1). [In Signa notation] i=1 It we want to calculate the sun than 3/22-1) n.(n+1) = 2 = 2 7 = n²+x-x an 1 IMS IMS i=1 1 + Ibn Ľ : 1 = 1 + Žbon no n=1 1 1-b it Ipl < 1 . Sum will be 1 + 1 / + 3 n=3 5 Ο Σ (n+1) 720 Let n+1=u then it n=0 then u=1 and if n=5 than u=6 6 6 5 E L So, Inti) 3r 2 n=1 u=1 no