Question

已知数列{a}的通项公式为an=48/[(n+2)^2-4],Sn是它的前n项和，则与S98最接近的整数是20,21,24还是25？

Answer 1

an=48/[(n+2)^2-4]
=48/[(n+2+2)(n+2-2)]
=48/(n*(n+4))
=12*[1/n-1/(n+4)]

Sn=12*(1-1/5+1/2-1/6+1/3-1/7+1/4-1/8+1/5-1/9+...+1/(n-4)-1/n+1/(n-3)-1/(n+1)+1/(n-2)-1/(n+2)+1/(n-1)-1/(n+3)+1/n-1/(n+4))

Sn=12*[1+1/2+1/3+1/4-1/(n+1)-1/(n+2)-1/(n+3)-1/(n+4)]
=12+6+4+3-12*[1/(n+1)+1/(n+2)+1/(n+3)+1/(n+4)]
=25-12*[1/(n+1)+1/(n+2)+1/(n+3)+1/(n+4)]

S98=25-12*(1/99+1/100+1/101+1/102)
>25-12*(1/99+1/99+1/99+1/99)
=25-48/99
>25-48/96
=24.5

S98>24.5

S98最接近的整数是 25