(1) 1+(-2)+3+(-4)+````````+99+(-100)； (2)1+(-2)+3+(-4)+5+(-6)+``````````+(2n-1)=(-2n)(1) 1+(-2)+3+(-4)+````````+99+(-100)=(2)1+(-2)+3+(-4)+5+(-6)+``````````+(2n-1)=(-2n)=这两题目等于多少？
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(1) 1+(-2)+3+(-4)+````````+99+(-100)； (2)1+(-2)+3+(-4)+5+(-6)+``````````+(2n-1)=(-2n)(1) 1+(-2)+3+(-4)+````````+99+(-100)=(2)1+(-2)+3+(-4)+5+(-6)+``````````+(2n-1)=(-2n)=这两题目等于多少？
(1) 1+(-2)+3+(-4)+````````+99+(-100)； (2)1+(-2)+3+(-4)+5+(-6)+``````````+(2n-1)=(-2n)
(1) 1+(-2)+3+(-4)+````````+99+(-100)=
(2)1+(-2)+3+(-4)+5+(-6)+``````````+(2n-1)=(-2n)=
这两题目等于多少？
(1) 1+(-2)+3+(-4)+````````+99+(-100)； (2)1+(-2)+3+(-4)+5+(-6)+``````````+(2n-1)=(-2n)(1) 1+(-2)+3+(-4)+````````+99+(-100)=(2)1+(-2)+3+(-4)+5+(-6)+``````````+(2n-1)=(-2n)=这两题目等于多少？
(1)、看做2个等差数列,一个首项为1,公差为2；另一个首项为-2,公差为-2,由求和公式得
Sn=(1+99)*50/2+(-2+-100)*50/2=-50 也可以将其化简,1-2=-1 ,后面同理共有50对,得Sn=-50
（2）、跟第一问一样,用简便方法则是Sn=-1*n=-n
1+(-2)+3+(-4)+...+99+(-100)
=(-1)*50=-50
-50
(1) -50
(2) -n
1+(-2)+3+(-4)+````````+99+(-100)=(1+3+5+7+9.+11+……99)-（2+4+6+8+10+…+100）=2500-2400-150=-50
1+(-2)+3+(-4)+5+(-6)+``````````+(2n-1)=(-2n)=（1+3+5+7+9+……+2n-1）-（2+4+6+…+2n）=n*n-n*n-n=-n
-50