已知对于任意正整数n都有a1+a2+...+an=n^3,则(1/a2-1)+(1/a3-1)+...+(1/a100-1)=_____
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![已知对于任意正整数n都有a1+a2+...+an=n^3,则(1/a2-1)+(1/a3-1)+...+(1/a100-1)=_____](/uploads/image/z/7837550-62-0.jpg?t=%E5%B7%B2%E7%9F%A5%E5%AF%B9%E4%BA%8E%E4%BB%BB%E6%84%8F%E6%AD%A3%E6%95%B4%E6%95%B0n%E9%83%BD%E6%9C%89a1%2Ba2%2B...%2Ban%3Dn%5E3%2C%E5%88%99%281%2Fa2-1%29%2B%281%2Fa3-1%29%2B...%2B%281%2Fa100-1%29%3D_____)
已知对于任意正整数n都有a1+a2+...+an=n^3,则(1/a2-1)+(1/a3-1)+...+(1/a100-1)=_____
已知对于任意正整数n都有a1+a2+...+an=n^3,则(1/a2-1)+(1/a3-1)+...+(1/a100-1)=_____
已知对于任意正整数n都有a1+a2+...+an=n^3,则(1/a2-1)+(1/a3-1)+...+(1/a100-1)=_____
a1+a2+...+a(n-1)+an=n³ (1)
a1+a2+...+a(n-1)=(n-1)³ (2)
(1)-(2)
an=n³-(n-1)³
=[n-(n-1)][n²+n(n-1)+(n-1)²]
=3n²-3n+1
1/(an -1)=1/(3n²-3n+1-1)=1/(3n²-3n)=(1/3)×1/(n²-n)=(1/3)×1/[n(n-1)]=(1/3)[1/(n-1)-1/n]
1/(a2-1)+1/(a3-1)+...+1/(a100 -1)
=(1/3)[1/1 -1/2+1/2-1/3+...+1/(99)-1/100]
=(1/3)(1 -1/100)
=(1/3)(99)/100
=33/100
Sn=n^3 (1)
S(n-1) =(n-1)^3 (2)
(1)-(2)
an= 3n^2-3n+1
an-1= 3n(n-1)
1/(an-1) = (1/3)( 1/(n-1) -1/n )
(1/(a2-1))+(1/(a3-1))+...+(1/(a100-1) )
= (1/3)( 1- 1/100)
= 33/100