数列An为等差数列,其前n项和为Sn 且a3=7,a5=11.求证1/s1+1/s2+.+1/Sn
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数列An为等差数列,其前n项和为Sn 且a3=7,a5=11.求证1/s1+1/s2+.+1/Sn<3/4急用!
![数列An为等差数列,其前n项和为Sn 且a3=7,a5=11.求证1/s1+1/s2+.+1/Sn](/uploads/image/z/17093982-30-2.jpg?t=%E6%95%B0%E5%88%97An%E4%B8%BA%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%2C%E5%85%B6%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn+%E4%B8%94a3%3D7%2Ca5%3D11.%E6%B1%82%E8%AF%811%2Fs1%2B1%2Fs2%2B.%2B1%2FSn)
设通项公式为an=a1+(n-1)d
11-7=a5-a3=(5-3)d
2d=4,
d=2
a3=a1+2d=a1+4=7
a1=3
Sn=n(n+2)
1/S1+1/S2+1/S3+……+1/Sn
=1/(1*3)+1/(2*4)+1/(3*5)+1/(4*6)+……+1/[n(n+2)]
=(1/2)[(1/1-1/3)+(1/2-1/4)+(1/3-1/5)+……+1/n-1/(n+2)]
=(1/2)[1/1+1/2-1/(n+1)-1/(n+2)]
11-7=a5-a3=(5-3)d
2d=4,
d=2
a3=a1+2d=a1+4=7
a1=3
Sn=n(n+2)
1/S1+1/S2+1/S3+……+1/Sn
=1/(1*3)+1/(2*4)+1/(3*5)+1/(4*6)+……+1/[n(n+2)]
=(1/2)[(1/1-1/3)+(1/2-1/4)+(1/3-1/5)+……+1/n-1/(n+2)]
=(1/2)[1/1+1/2-1/(n+1)-1/(n+2)]
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