Question

Write each arithmetic series in summation notation. 7+9+11+ . . . . . . +21

Answer 1

The given arithmetic series can be expressed in summation notation as $\sum\limits^8_{n=1} {(2n+5)}$ or in simplified notation $40+2 \sum\limits^8_{n=1} {n}$

The summation notation is used to express long summation in one expression.

The given arithmetic series is:

       7+9+11+ . . . . . . +21

Find the explicit formula for the nth term in the given series.

Recall:

a(n) = a(1) + (n-1) . d

where:

a(1) = the first term

d = common difference = a(2) - a(1) = a(n) - a(n-1)

From the series, we get:

a(1) = 7

d = 9 - 7 = 2

Hence,

a(n) =  a(1) + (n-1) . d

a(n) = 7 + (n-1) . 2

a(n) = 2n + 5

Find number of terms in the given series.

From the given series, we know that the last term is 21.

Let the last term be a(n), then,

a(n) = 2n + 5

21 = 2n + 5

2n = 16

n = 8

Write the summation notation with lower limit n = 1 and upper limit n = 8.

$\sum\limits^8_{n=1} {a(n)}= \sum\limits^8_{n=1} {(2n+5)}$

If necessary, simplify the summation.

$\sum\limits^8_{n=1} {(2n+5)}= \sum\limits^8_{n=1} {2n}+ \sum\limits^8_{n=1} {5}$

$= \sum\limits^8_{n=1} {2n}+ 8\times5$

$=40+2 \sum\limits^8_{n=1} {n}$

Learn more about how to write summation notation here:

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