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

Feb 25, 2017

${\sum}_{i = 1}^{9} \frac{1}{3 i}$

Explanation:

The first term is:

${u}_{1} = \frac{1}{3 \cdot 1}$

The second term is:

${u}_{2} = \frac{1}{3 \cdot 2}$

The third term is:

${u}_{2} = \frac{1}{3 \cdot 3}$

$\vdots$

The ${i}^{t h}$ term is:

${u}_{i} = \frac{1}{3 \cdot i}$

There are nine terms; so the sum is:

${\sum}_{i = 1}^{9} \frac{1}{3 i}$

Feb 25, 2017

${\sum}_{i = 1}^{9} \frac{1}{3 i}$

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, $\frac{1}{3}$ is multiplied by $\frac{1}{i}$ between each term. Since the sum goes from $i = 1$ until $i = 9$, it can be written in sigma notation as:

${\sum}_{i = 1}^{9} \frac{1}{3 i}$

Note: Since $\frac{1}{3}$ is common to each term, it could be factored out from each term, and the sigma notation could also be written as:

$\frac{1}{3} {\sum}_{i = 1}^{9} \frac{1}{i}$