化简：1+n的二次方分之一+（n+1）的二次方分之1,并把结果写成一个式子的平方 请数学大神帮忙啊

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数学之美团为你解答
1 + 1/n² + 1/(n+1)²
= [ n²(n+1)² + (n+1)² + n ² ] / [ n²(n+1)² ]
= [ n²(n+1)² + (2n²+2n)+1 ] / [ n²(n+1)²]
= [ n²(n+1)² + 2n(n+1) + 1] / [ n²(n+1)² ]
= [ n(n+1) + 1] ² / [ n²(n+1)² ]
= { 1+ 1 / [ n(n+1) ] } ²
或
1 + 1/n² + 1/(n+1)²
= 1 + [ n²+(n+1)²] / [n²(n+1)²]
= 1 + (2n²+2n+1) / [ n²(n+1)² ]
= 1 + [ 2n(n+1) + 1 ] / [ n²(n+1)² ]
= 1 + 2/[ n(n+1)] + 1/[ n(n+1) ]²
= { 1 + 1 / [ n (n+1) ] } ²