作业帮1/6,1/24,1/60,1/120,------,求通项,并求出前n项和
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1/6,1/24,1/60,1/120,------,求通项,并求出前n项和
1/6,1/24,1/60,1/120,------,求通项,并求出前n项和
1/6,1/24,1/60,1/120,------,求通项,并求出前n项和
通项an = 1/[n(n+1)(n+2)]
an = 1/n * [1/(n+1) - 1/(n+2)]
= 1/2 * [1/n - 1/(n+1) - 1/(n+1) + 1/(n+2)]
前n项和Sn = 1/2 * [(1 - 1/2 - 1/2 + 1/3) + (1/2 - 1/3 - 1/3 + 1/4) + (1/3 - 1/4 - 1/4 + 1/5) + ...+ (1/(n-1) - 1/n - 1/n + 1/(n+1)) + (1/n - 1/(n+1) - 1/(n+1) + 1/(n+2))]
= 1/2 * [(1 - 1/2 - 1/(n+1) + 1/(n+2)]
= 1/4 - 1/[2(n+1)(n+2)]
1/(n+2)(n+1)n,和你自己算吧
6=1*2*3
24=2*3*4
60=3*4*5
120=4*5*6
通项是an = 1/[n(n+1)(n+2)]
可以变形为an = 1/n * [1/(n+1) - 1/(n+2)]
= 1/n * 1/(n+1) - 1/n * 1/(n+2)
= 1/n - 1/(n+1) - 1/2 * 1/n + 1/2 * 1/(n+2...
全部展开
6=1*2*3
24=2*3*4
60=3*4*5
120=4*5*6
通项是an = 1/[n(n+1)(n+2)]
可以变形为an = 1/n * [1/(n+1) - 1/(n+2)]
= 1/n * 1/(n+1) - 1/n * 1/(n+2)
= 1/n - 1/(n+1) - 1/2 * 1/n + 1/2 * 1/(n+2)
= 1/2 * [1/n - 1/(n+1) - 1/(n+1) + 1/(n+2)]
观察Sn，Sn = 1/2 * [(1 - 1/2 - 1/2 + 1/3) + (1/2 - 1/3 - 1/3 + 1/4) + (1/3 - 1/4 - 1/4 + 1/5) + ... + (1/(n-1) - 1/n - 1/n + 1/(n+1)) + (1/n - 1/(n+1) - 1/(n+1) + 1/(n+2))]
相互抵消之后，只剩Sn = 1/2 * [(1 - 1/2 - 1/(n+1) + 1/(n+2)]
= 1/4 - 1/[2(n+1)(n+2)]
通项公式为an = 1/[n(n+1)(n+2)]
前n项和是Sn = 1/4 - 1/[2(n+1)(n+2)]
收起
An = 1/[n*(n+1)*(n+2)]
2An = [1/n - 1/(n+1)] - [1/(n+1) - 1/(n+2)]
其实这就是3项的裂项相消，2项的会做（如1/[n*(n+1)]），3项的也一样做
所以按这个方法得：
Sn = 1/2 - 1/(n+1) + 1/(n+2)
1/[n*(n+1)*(n+2)]
Sn我还没算出来