求解一道初三数学题，望高人指点，详细看
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(n+1)²+n²+n²(n+1)²
=(n+1)²-2n(n+1)+n²+2n(n+1)+n²(n+1)²
=[(n+1)-n]²+2n(n+1)+n²(n+1)²
=1+2n(n+1)+n²(n+1)²
=(n²+n+1)²
1+1/n²+1/(n+1)²
=[(n+1)²+n²+n²(n+1)²]/n²(n+1)²
=(n²+n+1)²/(n²+n)²
所以√[1+1/n²+1/(n+1)²]
=(n²+n+1)/(n²+n)
=1+1/(n(n+1)
=1+1/n-1/(n+1)
所以A=1+1-1/2+1+1/2-1/3+……+1+1/2005-1/2006
=1*2005+(1-1/2006)
=2005+2005/2006
接近2006
=(n+1)²-2n(n+1)+n²+2n(n+1)+n²(n+1)²
=[(n+1)-n]²+2n(n+1)+n²(n+1)²
=1+2n(n+1)+n²(n+1)²
=(n²+n+1)²
1+1/n²+1/(n+1)²
=[(n+1)²+n²+n²(n+1)²]/n²(n+1)²
=(n²+n+1)²/(n²+n)²
所以√[1+1/n²+1/(n+1)²]
=(n²+n+1)/(n²+n)
=1+1/(n(n+1)
=1+1/n-1/(n+1)
所以A=1+1-1/2+1+1/2-1/3+……+1+1/2005-1/2006
=1*2005+(1-1/2006)
=2005+2005/2006
接近2006
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