等差数列an中,a7=4,a19=2a9,求an的通项公式;设bn=1/nan,求数列bn前n项的和
2013-11-07
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a7=a1+6d=4;
a19=a1+18d=2(a1+8d);
a1=2d;
8d=4;
d=1/2;
a1=1;
∴an=a1+(n-1)d=1+(n-1)/2=(n+1)/2;
(2)bn=1/n(n+1)/2=2/n(n+1)=2×[1/n-1/(n+1)];
∴Sn=b1+b2+..+bn
=2×[1-1/2+1/2-1/3+...+1/n-1/(n+1)]
=2×n/(n+1)
=2n/(n+1);
a19=a1+18d=2(a1+8d);
a1=2d;
8d=4;
d=1/2;
a1=1;
∴an=a1+(n-1)d=1+(n-1)/2=(n+1)/2;
(2)bn=1/n(n+1)/2=2/n(n+1)=2×[1/n-1/(n+1)];
∴Sn=b1+b2+..+bn
=2×[1-1/2+1/2-1/3+...+1/n-1/(n+1)]
=2×n/(n+1)
=2n/(n+1);
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