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Compute  \(1 - 2 + 3 - 4 + \dots + 2005 - 2006 + 2007\)

 Jul 9, 2018
 #1
avatar+17771 
0

This isn't a geometric series ...

 

(1 - 2) + (3 - 4) + (5 - 6) + ... + (2005 - 2006) + 2007

   -1     +     -1   +     -1   + ... +          -1           + 2007

There are 1003 groups that result in - 1, so:   (1003)(-1) + 2007  =  -1003 + 2007  =  1004

 Jul 9, 2018
 #2
avatar
0

You could also use arithmetic sequence formula to sum them up as two separate sequences:

 

1 + 3 + 5 + 7 +...............+ 2007 =1,008,016.00

-2 - 4 - 6 - 8 - ................- 2006 =-1,007,012.00

1,008,016.00 + (-1,007,012.00) =1,004

 Jul 9, 2018
 #3
avatar+22182 
0

Compute:

\(1 - 2 + 3 - 4 + \dots + 2005 - 2006 + 2007\)

 

\(\small{ \begin{array}{|r|rcrcrcrcccrcrcrl|} \hline \text{sum }= & 1 &-& 2 &+& 3 &-& 4 &+& \dots &+& 2005 &-& 2006 &+& 2007 \\ \text{sum }= & 2007 &-& 2006 &+& 2005 &-& 2004 &+& \dots &+& 3 &-& 2 &+& 1 \\ \hline 2\cdot \text{sum } = & 2008 &-& 2008 &+& 2008 &-& 2008 &+& \dots &+& 2008 &-& 2008 &+& 2008 \\ 2\cdot \text{sum } = & & & 0 &+& && 0 &+& \dots &+& && 0 &+& 2008 \\ 2\cdot \text{sum } = & & & & & && & & & & && & & 2008 & \quad | \quad : 2 \\ \text{sum } = & & & & & && & & & & && & & \mathbf{1004} \\ \hline \end{array} } \)

 

laugh

 Jul 9, 2018

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