1.已知：a的平方-3a+1=0,求a的平方/a的四次方+1的值.2.计算：[1].1 1/2（一又二分之一）+3 1/4+5 1/8+7 1/16+...+17 1/512;[2].1/2+3/4+5/8+7/16+...+19/1024;[3].1/1x4+1/4x7+1/7x10+...+1/(3n-2)x(3n+1);[4]1x2+2x3+3x4+...+n(n+1).♀¤

1.已知：a的平方-3a+1=0,求a的平方/a的四次方+1的值.2.计算：[1].1 1/2（一又二分之一）+3 1/4+5 1/8+7 1/16+...+17 1/512;[2].1/2+3/4+5/8+7/16+...+19/1024;[3].1/1x4+1/4x7+1/7x10+...+1/(3n-2)x(3n+1);[4]1x2+2x3+3x4+...+n(n+1).♀¤
1.已知：a的平方-3a+1=0,求a的平方/a的四次方+1的值.
2.计算：
[1].1 1/2（一又二分之一）+3 1/4+5 1/8+7 1/16+...+17 1/512;
[2].1/2+3/4+5/8+7/16+...+19/1024;
[3].1/1x4+1/4x7+1/7x10+...+1/(3n-2)x(3n+1);
[4]1x2+2x3+3x4+...+n(n+1).
♀¤请于2008.8.7 12：00前做完¤♀

1.已知：a的平方-3a+1=0,求a的平方/a的四次方+1的值.2.计算：[1].1 1/2（一又二分之一）+3 1/4+5 1/8+7 1/16+...+17 1/512;[2].1/2+3/4+5/8+7/16+...+19/1024;[3].1/1x4+1/4x7+1/7x10+...+1/(3n-2)x(3n+1);[4]1x2+2x3+3x4+...+n(n+1).♀¤
1.因为a^2-3a+1=0,所以a+1/a=3,
所以(a+1/a)^2=a^2+1/a^2+2=9,
所以a^2+1/a^2=7,
所以a^2/(a^4+1)=1/(a^2+1/a^2)=1/7；
2.[1]原式=(1+3+5+...+17)+(1/2+1/4+...+1/512)
=9^2+1/2*[1-(1/2)^9]/(1-1/2)
=81+511/512
=81 511/512；
[2]因为通项公式是an=(2n-1)/2^n,
所以Sn=a1+a2+...+an
=1/2+3/4+...+(2n-1)/2^n,
所以Sn/2=1/4+3/8+...+(2n-3)/2^n+(2n-1)/2^(n+1),
所以Sn-Sn/2=1/2+2/4+2/8+...+2/2^n-(2n-1)/2^(n+1),
所以Sn/2=1/2+1-1/2^(n-1)-(2n-1)/2^(n+1)
=3/2-(2n+3)/2^(n+1),
所以Sn=3-(2n+3)/2^n,
所以1/2+3/4+5/8+7/16+...+19/1024
=S10=3-23/1024=2 1001/1024；
[3]因为1/(3n-2)(3n+1)=[1/(3n-2)-1/(3n+1)]/3,
所以1/1x4+1/4x7+1/7x10+...+1/(3n-2)x(3n+1)
=[1/1-1/4+1/4-1/7+...+1/(3n-2)-1/(3n+1)]/3
=n/(3n+1)；
[4]因为1x2+2x3+3x4+...+n(n+1)
=(1^2+2^2+...+n^2)+(1+2+...+n)
=n(n+1)(2n+1)/6+n(n+1)/2
=n(n+1)(n+2)/3.

1.有已知得a^2=3a-1
a^2/(a^4+1)=(3a-1)/[(3a-1)^2+1]
=(3a-1)/(9*a^2-6a+2)
=(3a-1)/[9(3a-1)-6a+2]
=1/7
2.原式=(1+1/2)+(3+1/4)+(5+1/8)+...+(17+1/512)
=(1+3+5+...+17)+(1/2+1/4+1/8+....

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1.有已知得a^2=3a-1
a^2/(a^4+1)=(3a-1)/[(3a-1)^2+1]
=(3a-1)/(9*a^2-6a+2)
=(3a-1)/[9(3a-1)-6a+2]
=1/7
2.原式=(1+1/2)+(3+1/4)+(5+1/8)+...+(17+1/512)
=(1+3+5+...+17)+(1/2+1/4+1/8+...+1/512)
=81+1023/1024
3.错位相减原式=S=1/2+3/4+5/8+7/16+...+19/1024则
1/2*S= 1/4+3/8+5/16+...+17/1024+19/2048
将上面两个式子做差得 1/2*S=1/2+2/4+2/8+2/16+...+2/1024-19/2048
=1/2+2*(1/4+1/8+...+1/1024)-19/2048
(通向公式是等差*等比，求和时用错位相减)
4.裂项相消原式=1/3[(1-1/4)+(1/4-1/7)+(1/7-1/10)+...+1/(3n-2)-1/(3n+1)]
=1/3[1-1/(3n+1)]
5.n*（n+1）=n^2+n,
原式=(1^2+1)+(2^2+2)+(3^2+3)+...+(n^2+n)
=(1^2+2^2+...+n^2)+(1+2+3+...+n)
=[n*(n+1)*(2n+1)]/6+[n*(n+1)]/2
=[n(n+1)(n+2)]/3

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