1/n(n+1)(n+2)求和怎么做

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1/n(n+1)(n+2)求和怎么做
1/n(n+1)(n+2)求和怎么做

1/n(n+1)(n+2)求和怎么做
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)[(n+2)/n(n+1)(n+2)-n/n(n+1)(n+2)]
=(1/2)[1/n(n+1)-1/(n+1)(n+2)]
所以和=(1/2)[1/1*2-1/2*3+1/2*3-1/3*4+……+1/n(n+1)-1/(n+1)(n+2)]
=(1/2)[1/1*2-1/(n+1)(n+2)]
=n(n+3)/[4(n+1)(n+2)]

1/n(n+1)(n+2)=1/2n-1/(n+1)+1/2(n+2)
用累加法：
a1=1/2*1-1/2+1/2*3
a2=1/2*2-1/3+1/2*4
a3=1/2*3-1/4+1/2*5
…………………………
an=1/2n-1/(n+1)+1/2(n+2)
Sn=1/4+1/2(n+1)-1/(n+1)+1/2*(n+2)
=1/4-1/2(n+1)(n+2)

如图

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