1/2*4+1/4*6+1/6*8...+1/2n(2n+2)=1001/4008 最好有解析
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1/2*4+1/4*6+1/6*8...+1/2n(2n+2)=1001/4008 最好有解析
1/2*4+1/4*6+1/6*8...+1/2n(2n+2)=1001/4008 最好有解析
1/2*4+1/4*6+1/6*8...+1/2n(2n+2)=1001/4008 最好有解析
列项求和
1/[2n*(2n+2)]
=1/2*[1/(2n)-1/(2n+2)]
原式化为：
1/2*[1/2-1/4+1/4-1/6+1/6-1/8+.+1/(2n)-1/(2n+2)]=1001/4008
1/2-1/(2n+2)=1001/4008*2
1/2-1/(2n+2)=1001/2004
1/(2n+2)=1/2-1001/2004
1/(2n+2)=1/2004
2n+2=2004
2n=2002
n=1001
1/2n(2n+2)
=1/2*2/2n(2n+2)
=1/2*[(2n+2)-2n]/2n(2n+2)
=1/2*[(2n+2)/2n(2n+2)-2n/2n(2n+2)]
=1/2*[1/2n-1/(2n+2)]
所以1/2*[1/2-1/4+1/4-1/6+……+1/2n-1/(2n+2)]=1001/4008
1/2-1/(2n+2)]=1001/2004
1/(2n+2)=1/2-1001/2004=1/2004
2n+2=2004
n=1001
1/2n(2n+2)=1/2*(1/n(n+1))=1/2*(1/n-1/(n+1))
所以最后答案是1/2*（1-1/(n+1））
1/2*4=1/2(1/2-1/4)
1/4*6=1/2(1/4-1/6)
...
1/2n(2n+2)=1/2(1/2n-1/2n+2)
...
所以左边=1/2(1/2-1/4+1/4-1/6+...+1/2n-1/2n+2)
=1/2(1/2-1/2n+2)
=1/4*n(n+1)
右边=1001/4008
所以解得n=1001
1/2*4+1/4*6+1/6*8...+1/2n(2n+2)
=1/4[1/1*2+1/2*3+1/3*4...+1/n(n+1)]
=1/4[2/3+1/3*4...+1/n(n+1)]
=1/4[3/4+...+1/n(n+1)]
=1/4[n/(n+1)]
=1/4(1001/1002)
所以n=1001
高中题？好像有一种叫分母分裂求和法的。自己去搞定哈~~