Question 809009: The sum
1 - 2 + 3 - 4 + ... + 65 - 66 + 67
is
Answer by mananth(16946) (Show Source):
You can put this solution on YOUR website! 1 - 2 + 3 - 4 + ... + 65 - 66 + 67
let us break it up into two sequences
both are arithmetic progressions
1+3+5..........67
a=1,d=2 Tn=67
67=1+(n-1)2
67=1+2n-2
2n=68
n=34
Sn = n/2(2a+(n-1)d)
=(34/2(2*1+(37-1)2)
=17*(2+36*2)
=17*74
=1258
and -2,-4,-6......-66
a=-2, d=-2 Tn= -66
-66=-2+(n-1)*-2
-66=-2-2n+2
n=33
S33= 33/2( 2*-2+(33-1)*-2)
S33=33/2(-4-64)
S33=33/2 * -68
=33*-34
=-1122
1258-1122
=136
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