题目内容
已知函数
且an=f(n)+f(n+1),
则a
+a
+a
+…+a
等于 .
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232125213642887.jpg)
则a
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823212521379195.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823212521395240.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823212521410240.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823212521426283.png)
100
解:当n为奇数时,an=f(n)+f(n+1)=n2-(n+1)2=-2n-1,
当n为偶数时,an=f(n)+f(n+1)=-n2+(n+1)2=2n+1,
则S100=(a1+a3+a5+a7+..+a99)+(a2+a4+a6+a8+….+a100)
=-2×(1+3+5+77+..+99)-5+2×(2+4+6+8++…+100)+5
=100
当n为偶数时,an=f(n)+f(n+1)=-n2+(n+1)2=2n+1,
则S100=(a1+a3+a5+a7+..+a99)+(a2+a4+a6+a8+….+a100)
=-2×(1+3+5+77+..+99)-5+2×(2+4+6+8++…+100)+5
=100
![](http://thumb2018.1010pic.com/images/loading.gif)
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