+0  
 
0
81
2
avatar

How do you compute this sum?

 

\( \large1-2+3+4-5+6+7-8+9+10-\cdots-2000+2001+2002-2003 =\, ?\)

 Jul 8, 2020
 #1
avatar+21955 
+2

If the sum were:  (1 - 2 + 3) + (4 - 5 + 6) + (7 - 8 + 9) + ... + (1999 - 2000 + 2001) + (2002 - 2003 + 2004)

we could do this:       2         +       5        +        8        + ... +             2000               +            2003

and we would have an arithmetic series with:  t1 = 2     tn = 2003     d = 3.

 

We could then use this formula;  tn  =  t1 + (n - 1)d  to find the number of terms:

     --->     2003  =  2 + (n - 1)3     --->     2001  =  (n - 1)3     --->     667  =  n - 1     --->   n  =  668  

 

and use this formula:  Sum  =  n(t1 + tn)/2  to find the sum:

     --->     Sum  =  668(2 + 2003)/2     --->     Sum  =  669 670

 

However, we don't have  2004  in the original problem, so we'll have to subtract that:

     --->     669 670 - 2004  =  667 666

 Jul 8, 2020
 #2
avatar
0

sumfor(n, 1, 667, (3*n - 1) =667,667 + 2002 - 2003 =667,666

 Jul 8, 2020

52 Online Users

avatar
avatar