How do you use sigma notation to write the sum for #1/(3(1))+1/(3(2))+1/(3(3))+...+1/(3(9))#?

2 Answers
Feb 25, 2017

Answer:

# sum_(i=1)^9 1/(3i) #

Explanation:

The first term is:

# u_1 = 1/(3*1) #

The second term is:

# u_2 = 1/(3*2) #

The third term is:

# u_2 = 1/(3*3) #

#vdots#

The #i^(th)# term is:

# u_i = 1/(3*i) #

There are nine terms; so the sum is:

# sum_(i=1)^9 1/(3i) #

Feb 25, 2017

Answer:

#sum_(i=1)^(9)1/(3i)#

Explanation:

Sigma notation is used in mathematics to condense a sum of terms into a more readable format. Each term will vary according to an integer, i, which is incremented by 1. (i.e. #i# increases by 1 for each term.)

In sigma notation, the term below the #Sigma# defines the starting point for #i#, and the term on top of the #Sigma# defines the stopping point for #i#. For example:

#sum_(i=0)^(3)i=0+1+2+3=6#

In this example, #i# starts at 0 and ends at 3. #i# is incremented by +1 for each term (0, 1, 2, 3), resulting in a total sum of 6.

Looking at the original question, #1/3# is multiplied by #1/i# between each term. Since the sum goes from #i=1# until #i=9#, it can be written in sigma notation as:

#sum_(i=1)^(9)1/(3i)#

Note: Since #1/3# is common to each term, it could be factored out from each term, and the sigma notation could also be written as:

#1/3sum_(i=1)^(9)1/i#