${(- 1)}^{- 10} + {(- 1)}^{- 9} + {(- 1)}^{- 8} + \dots + {(- 1)}^{9} + {(- 1)}^{10}$ $(-1)^{-10} + (-1)^{-9} + (-1)^{-8} + \cdots + (-1)^9 + (-1)^{10}$ (The dots $\dots$ $\cdots$ mean that there are 21 numbers being added, one for each integer from $- 10$ $-10$ to 10.) ?

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${(- 1)}^{- 10} + {(- 1)}^{- 9} + {(- 1)}^{- 8} + \dots + {(- 1)}^{9} + {(- 1)}^{10}$ $(-1)^{-10} + (-1)^{-9} + (-1)^{-8} + \cdots + (-1)^9 + (-1)^{10}$ (The dots $\dots$ $\cdots$ mean that there are 21 numbers being added, one for each integer from $- 10$ $-10$ to 10.) ?

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Answer 1 · 2018-07-23T22:45:55

The sum of the sequence is $1$ $1$

Explanation:

Logically, if we are adding $1$ $1$ and $- 1$ $-1$ repeatedly, the sum is $0$ $0$ , but since the first and last terms of the sequence are both $1$ $1$ , we know that in the sequence there is one more $1$ $1$ than $- 1$ $-1$ .

We can prove it with a geometric sum formula for finite sums:
$S_{n} = a_{1} (\frac{1 - r^{n}}{1 - r})$ $S_n= a_1((1-r^n)/(1-r))$

$S_{21} = 1 (\frac{1 - {(- 1)}^{21}}{1 - (- 1)})$ $S_21= 1((1-(-1)^(21))/(1-(-1)))$

$S_{21} = 1 (\frac{2}{2})$ $S_21= 1(2/2)$

$S_{21} = 1$ $S_21= 1$

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${(- 1)}^{- 10} + {(- 1)}^{- 9} + {(- 1)}^{- 8} + \dots + {(- 1)}^{9} + {(- 1)}^{10}$ $(-1)^{-10} + (-1)^{-9} + (-1)^{-8} + \cdots + (-1)^9 + (-1)^{10}$ (The dots $\dots$ $\cdots$ mean that there are 21 numbers being added, one for each integer from $- 10$ $-10$ to 10.) ?

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