(1) 1+(-2)+3+(-4)+````````+99+(-100); (2)1+(-2)+3+(-4)+5+(-6)+``````````+(2n-1)=(-2n)
问题描述:
(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?
答
(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