请教高一数学,(题目中的分子都为1),请列步骤。谢谢
6个回答
展开全部
1/[(n+1)²-1]=1/(n²+2n)=1/[n(n+2)]=(1/2)*[(1/n)-(1/n+2)]
所以:
原式=(1/2)*[1-(1/3)]+(1/2)*[(1/2)-(1/4)]+(1/2)*[(1/3)-(1/5)]+……+(1/2)*[(1/n-2)-(1/n)]+(1/2)*[(1/n-1)-(1/n+1)]+(1/2)*[(1/n)-(1/n+2)]
=(1/2)*[1-(1/3)+(1/2)-(1/4)+(1/3)-(1/5)+……+(1/n-2)-(1/n)+(1/n-1)-(1/n+1)+(1/n)-(1/n+2)]
=(1/2)*[1+(1/2)-(1/n+1)-(1/n+2)]
=(3/4)-(1/2)*[(2n+3)/(n+1)(n+2)]
所以:
原式=(1/2)*[1-(1/3)]+(1/2)*[(1/2)-(1/4)]+(1/2)*[(1/3)-(1/5)]+……+(1/2)*[(1/n-2)-(1/n)]+(1/2)*[(1/n-1)-(1/n+1)]+(1/2)*[(1/n)-(1/n+2)]
=(1/2)*[1-(1/3)+(1/2)-(1/4)+(1/3)-(1/5)+……+(1/n-2)-(1/n)+(1/n-1)-(1/n+1)+(1/n)-(1/n+2)]
=(1/2)*[1+(1/2)-(1/n+1)-(1/n+2)]
=(3/4)-(1/2)*[(2n+3)/(n+1)(n+2)]
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展开全部
(n+1)^2-1=(n+1+1)(n+1-1)=n(n+2)
1/[n(n+2)]=1/2 [1/n- 1/(n+2)]
每个都这么代进去,把1/2提出来:
1/2 [(1/2-1/4)+(1/3-1/5)+(1/4-1/6)+……+(1/(n-1)-1/(n+1))+(1/n-1/(n+2))+(1/(n+1)-1(n+3))]
注意,中间很多项正好一正一负,都可以消去,只剩下头两项正的后两项负的
1/2 [ 1/2+1/3-1/(n+2)-1/(n+3)]
1/[n(n+2)]=1/2 [1/n- 1/(n+2)]
每个都这么代进去,把1/2提出来:
1/2 [(1/2-1/4)+(1/3-1/5)+(1/4-1/6)+……+(1/(n-1)-1/(n+1))+(1/n-1/(n+2))+(1/(n+1)-1(n+3))]
注意,中间很多项正好一正一负,都可以消去,只剩下头两项正的后两项负的
1/2 [ 1/2+1/3-1/(n+2)-1/(n+3)]
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