数列难死a1=-5/12 an=a(n-1)+1/(n^+2n)an=?

an=a(n-1)+1/2x(1/n - 1/n+2)
a(n-1)=a(n-2)+1/2x(1/n-1 - 1/n+1)
a(n-2)=a(n-3)+1/2x(1/n-2 - 1/n )
a(n-3)=a(n-4)+1/2x(1/n-3 - 1/n-1)
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.
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a4=a3 +1/2x(1/4 - 1/6)
a3=a2 +1/2x(1/3 - 1/5)
a2=a1 +1/2x(1/2 - 1/4)
全部等式对应相减,得
an=a1+1/2x( 1/2 +1/3 - 1/n+1 - 1/n+2)
=5/6 -1/2x( 1/n+1 + 1/n+2)

裂项 1/(n^+2n) =1/2{1/n-1/(n+2)} 所以
an-a(n-1)=1/2{1/n-1/(n+2)}
迭代，就可以了

应用1/(n^+2n)=1/2*(1/n-1/n+2）即可