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设S1=1+1/1?+1/2?,S2=1+1/2?+1/3?.S=根号S1+根号S2.
题目详情
设S1=1+1/1?+1/2?,S2=1+1/2?+1/3?.S=根号S1+根号S2.
▼优质解答
答案和解析
Sn=1+1/n^2+1/(n+1)^2=(n^4+2n^3+3n^2+2n+1)/(n^2*(n+1)^2)=(n*(n+1)+1)^2/(n^2*(n+1)^2)
故√Sn=√(n*(n+1)+1)^2/(n^2*(n+1)^2)=[n(n+1)+1]/[n(n+1)]
所以:
√S1=1+1-1/2
√S2=1+1/2-1/3
√S3=1+1/3-1/4
.
√Sn=1+1/n-1/(n+1)
s= 1+1-1/2 +1+1/2-1/3 1+1/3-1/4 +1+1/(n(n+1)))=n+[(1-1/2)+(1/2-1/3)+...+(1/n-1/(n+1))]=n+1-1/(n+1)
故√Sn=√(n*(n+1)+1)^2/(n^2*(n+1)^2)=[n(n+1)+1]/[n(n+1)]
所以:
√S1=1+1-1/2
√S2=1+1/2-1/3
√S3=1+1/3-1/4
.
√Sn=1+1/n-1/(n+1)
s= 1+1-1/2 +1+1/2-1/3 1+1/3-1/4 +1+1/(n(n+1)))=n+[(1-1/2)+(1/2-1/3)+...+(1/n-1/(n+1))]=n+1-1/(n+1)
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