# How do I use summation notation to write the series 2.2 + 6.6?

Aug 11, 2016

$2.2 {\sum}_{i = 0 \to n - 1} {3}^{i}$

#### Explanation:

Just two terms is not really enough to build the series with confidence. However, lets build based on an assumption:

Let each term be represented by ${a}_{i}$ where $i = 1 , 2 , 3 , 4. . . n$

Let
${a}_{i} \to {a}_{1} = 2.2$
${a}_{i} \to {a}_{2} = 6.6$
${a}_{i} \to {a}_{3} = 19.8$
${a}_{i} \to {a}_{4} = 59.4$
and so on.

Let the sum be $s$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Notice that
${a}_{2} = 3 {a}_{1}$
${a}_{3} = 3 {a}_{2} = 3 \times 3 {a}_{1}$
${a}_{4} = 3 {a}_{3} = 3 \times 3 \times 3 {a}_{1}$

Notice that we have ${a}_{i} = {3}^{i - 1} \left(2.2\right)$

Test: ${a}_{1} = {3}^{1 - 1} \left(2.2\right) = {3}^{0} \left(2.2\right) = 1 \left(2.2\right) = 2.2$

$s = {a}_{1} + {a}_{2} + {a}_{3} + \ldots + {a}_{n}$

$s = {3}^{0} \left(2.2\right) + {3}^{1} \left(2.2\right) + \ldots + {3}^{i - 1} \left(2.2\right) + \ldots + {3}^{n - 1} \left(2.2\right)$

Factor out the 2.2 giving:

$2.2 \left({3}^{0} + {3}^{1} + {3}^{2} + \ldots + {3}^{i - 1} + \ldots {3}^{n - 1}\right)$

So the notation is

$2.2 {\sum}_{i = 0 \to n - 1} {3}^{i}$