Write each series using sigma notation. Find the sum. - 7 - 42 - 252 - ... - 54.432

Solution

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Recall the sigma notation:

\sum\limits_{n=i}^{j}a_n

where $n$ is the index of summation, $i$ is the lower limit of summation, $j$ is the upper limit of summation, and $a_n$ is the general term.

The series is geometric where $a_1=-7$ and $r=6$ so the general term is the explicit definition of the corresponding geometric sequence:

a_n=-7\left(6\right)^{n-1}

We let $i=1$ so that $a_{6}=-54432$ which follows that $j=6$ . So, the sigma notation for the geometric series is:

\color{#c34632}{\sum\limits_{n=1}^{6}-7\left(6\right)^{n-1} }

To find the sum, we use the formula for the sum of a finite geometric series given by:

S_n=\dfrac{a_1(1-r^n)}{(1-r)}

where $a_1$ is the first term and $r$ is the common ratio.

From the series, $n=6$ , $a_1=-7$ , and $r=6$ so the sum is:

S_6=\dfrac{-7\left[1-\left(6\right)^6\right]}{\left(1-6\right)}

S_6=\dfrac{-7\left(-46655\right) }{-5}

\color{#c34632}{S_6=-65317}

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