Answer
$-399$
Work Step by Step
See p. 855.
For the arithmetic sequence $a_{n}=a+(n-1)d$
the nth partial sum $S_{n}=\displaystyle \sum_{k=1}^{n}[a+(k-1)d]$
is given by either of the following equivalent formulas:
1. $S_{n}=\displaystyle \frac{n}{2}[2a+(n-1)d]\qquad $2. $S_{n}=n(\displaystyle \frac{a+a_{n}}{2})$
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$\displaystyle \sum_{n=20}^{20}(1-2n)$.
is a partial sum of an arithmetic series for which the first term is
$a= 1-2\cdot 0=1$
and $d=-2$
There are $21$ terms in the sum (for n=0,1,2,...,20)
The last term is $a_{21}=1-2\cdot 20=-39$.
Using formula 2,
$S_{21}=\displaystyle \frac{21}{2}(1-39)=-399$
