Question

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

Answer 1

`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)`

Answer 2

`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`