一道高一数学题,急,高手进,好的加分.

因式分解 n^4 + n^2 + 1

n^4 + n^2 + 1
= n^4 + 2n^2 - n^2 + 1
= (n^2 + 1)^2 - n^2
= (n^2 + 1 + n)(n^2 + 1 - n)

那么
a<n>
= n/(n^4 + n^2 + 1)
= n/[(n^2 + 1 + n)(n^2 + 1 - n)]
= n * [ 1/(n^2 + 1 - n) - 1/(n^2 + 1 + n) ]
= n * 1/(2n) * [ 1/(n^2 + 1 - n) - 1/(n^2 + 1 + n) ]
= 1/2 * [ 1/(n^2 + 1 - n) - 1/(n^2 + 1 + n) ]

a<n-1>
= 1/2 * [ 1/((n-1)^2 + 1 - n + 1) - 1/((n-1)^2 + 1 + n-1) ]
= 1/2 * [ 1/(n^2 - 2n + 1 + 1 - n + 1) - 1/(n^2-2n + 1 + 1 + n-1) ]
= 1/2 * [ 1/(n^2 - 3n + 3) - 1/(n^2 - n + 1) ]

所以
S<n>
= a<1> + a<2> + ... + a<n-1> + a<n>
= 1/2 * (1 - 1/(n^2 + 1 + n) )
= 1/2 * (n^2 + n)/(n^2 + 1 + n)

n^4+n^2+1
=n^4+2n^2+1-n^2
=(n^2+1)^2-n^2
=(n^2+n+1)(n^2-n+1)

n^2-n+1
=(n-1)^2+(n-1)+1

n/(n^4+n^2+1)
=[2n/(n^2+n+1)(n^2-n+1)]/2
=[1/(n^2-n+1)-1/(n^2+n+1)]/2
={1/[(n-1)^2+