1／n(n+1)(n+2)裂项求和转化

问题描述：

1／n(n+1)(n+2)裂项求和转化

答

先把n(n+1)看成一项不行吗

？图片没看到吗你把这数列题目拍下来吧1/[n(n+1)(n+2)] =1/2*{2/[n(n+1)(n+2)]}=1/2*{[(n+2)-n]/n(n+1)(n+2)}=1/2*{1/n(n+1) - 1/(n+1)(n+2)}证毕！？1/[n(n+1)(n+2)] =1/2*{2/[n(n+1)(n+2)]}=1/2*{[(n+2)-n]/n(n+1)(n+2)}=1/2*{1/n(n+1) - 1/(n+1)(n+2)} 看到了吗唔1/1*2*3+1/2*3*4+1/3*4*5+....+1/n(n+1)(n+2)+......
sn=1/2*[1/1*2-1/2*3+1/2*3-1/3*4+1/3*4-1/4*5+.....+1/n(n+1)-1/(n+1)(n+2)]
=1/2*[1/2-1/(n+1)(n+2)]
=1/4-1/2(n+1)(n+2)
limsn=1/4
所以
级数的和为1/4.实在看不懂就问老师去

1／n(n+1)(n+2)裂项求和转化

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