Question

A=48*（1/(3*3-4)+1/(4*4-4)+1/(5*5-4)+...+1/(100*100-4)）请问A最接近哪个整数？

我要具体解题过程哦！

Answer 1

A=48*(1/(3^2-4)+1/(4^2-4+~~~1/(100^2-4)
=48*[1/(1*5)+1/(2*6)+...+1/(98*102)]
=12*[(1/1-1/5)+(1/2-1/6)+...+(1/98-1/102)]
=12*[1+1/2+1/3+1/4-1/99-1/100-1/101-1/102]
=12+6+4+3-12/99-12/100-12/101-12/102
=25-(12/99+12/100+12/101+12/102).
∵0<12/102<12/101<12/100<12/99<1/8,
∴0<12/99+12/100+12/101+12/102<1/2,
与A最接近的整数是25.

Answer 2

A=48[1/(1*5)+1/(2*6)+1/(3*7)+1/(4*8)+1/(5*9)+...+1/(94*98)+1/(95*99)+1/(96*100)+1/(97*101)+1/(98*102)]1/5=（1/4）（1-1/5），1/12=(1/4)(1/2-1/6) ,1/21=(1/4)(1/3-1/7),以此类推然后提取1/4得到A=48*1/4*(1-1/5+1/2-1/6+1/3-1/7+1/4-1/8+1/5-1/9+...+1/94-1/98+1/95-1/99+1/96-1/100+1/97-1/101+1/98-1/102)有的可以约去得到A=12(1+1/2+1/3+1/4-1/99-1/100-1/101-1/102) =25-12(1/99+1/100+1/101+1/102)接近于25

Answer 3

1/(n^2-4)=1/(n-2)(n+2)=1/4(1/(n-2)-1/(n+2))所以A=12*（1-1/5+1/2-1/6+1/3-1/7+1/4-1/8+1/5-1/9+.....................+1/98-1/102）=12*(1+1/2+1/3+1/4-1/99-1/100-1/101-1/102)现在自己估算下不就知道了