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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
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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!
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答案和解析
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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