# How do you use sigma notation to write the sum for 3-9+27-81+243-729?

Feb 27, 2017

sum_(n=0)^5 [3*("–"3)^n]" " or $\text{ "-sum_(n=1)^6 ("–"3)^n" }$ or " "sum_(n=1)^6 ("–"1)^(n-1)3^n

#### Explanation:

In order to use summation notation, we need to find something that increments by 1 with each new term.

What we need to do is discern how each successive term differs from the previous one. Without looking too hard, we can see that each term is $\text{–} 3$ times the previous term. Thus, starting at our first term of 3, each new term is "3 times the next power of $\text{–} 3$":

$\text{ } 3 = 3$
"  –"9=3xx("-"3)^1
"  "27=3xx("-"3)^2
"–"81=3xx("-"3)^3

and so on.

What we notice is that the powers of -3 are going up by 1 with each new term. This is the increment value we will use in our sigma notation.

Our sum can be thought of as:

$3 - 9 + 27 - 81 + 243 - 729$
$= 3 + \left(\text{-"9)+27+("-"81)+243+("-} 729\right)$
$= \left[3 \cdot {\left(\text{-"3)^0]+[3*("-"3)^1]+[3*("-"3)^2]+[3*("-"3)^3]+[3*("-"3)^4]+[3*("-} 3\right)}^{5}\right]$

Which is written in sigma notation as:

${\sum}_{n = 0}^{5} \left[3 \cdot {\left(\text{–} 3\right)}^{n}\right]$

This says we wish to sum the expression $3 \cdot {\left(\text{–} 3\right)}^{n}$ a total of 6 times—once for each $n$ from $\left\{0 , 1 , 2 , 3 , 4 , 5\right\}$.

There are other ways to write the sum; they are included as extras in the Answer portion.