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17.
5-√9<5-√7<5-√4
2<5-√7<3,即5-√7在2到3之间
m=2,n=5-√7-2=3-√7
m=2,n=3-√7代入amn+bn²=1,得
a·2·(3-√7)+b·(3-√7)²=1
(4a+6b-3)-√7(2b+1)=0
a、b是有理数,要等式成立,只有
4a+6b-3=0
2b+1=0
解得a=3/2,b=-½
2a+b=2·(3/2)+(-½)=5/2
18.
Sn=1+ 1/n²+ 1/(n+1)²
=[n²(n+1)²+(n+1)²+n²]/[n²(n+1)²]
=[n²(n+1)²+n²+2n+1+n²]/[n(n+1)]²
=[n²(n+1)²+2n(n+1)+1]/[n(n+1)]²
=[n(n+1)+1]²/[n(n+1)]²
√Sn=√{[n(n+1)+1]²/[n(n+1)]²}=[n(n+1)+1]/[n(n+1)]=1+ 1/n -1/(n+1)
S=√S1+√S2+...+√Sn
=1+1/1 -1/2 +1+1/2 -1/3+...+1+ 1/n -1/(n+1)
=(1+1+...+1)+[1/1 -1/2 +1/2 -1/3+...+1/n -1/(n+1)]
=n+[1- 1/(n+1)]
=n+ n/(n+1)
=n(n+2)/(n+1)
5-√9<5-√7<5-√4
2<5-√7<3,即5-√7在2到3之间
m=2,n=5-√7-2=3-√7
m=2,n=3-√7代入amn+bn²=1,得
a·2·(3-√7)+b·(3-√7)²=1
(4a+6b-3)-√7(2b+1)=0
a、b是有理数,要等式成立,只有
4a+6b-3=0
2b+1=0
解得a=3/2,b=-½
2a+b=2·(3/2)+(-½)=5/2
18.
Sn=1+ 1/n²+ 1/(n+1)²
=[n²(n+1)²+(n+1)²+n²]/[n²(n+1)²]
=[n²(n+1)²+n²+2n+1+n²]/[n(n+1)]²
=[n²(n+1)²+2n(n+1)+1]/[n(n+1)]²
=[n(n+1)+1]²/[n(n+1)]²
√Sn=√{[n(n+1)+1]²/[n(n+1)]²}=[n(n+1)+1]/[n(n+1)]=1+ 1/n -1/(n+1)
S=√S1+√S2+...+√Sn
=1+1/1 -1/2 +1+1/2 -1/3+...+1+ 1/n -1/(n+1)
=(1+1+...+1)+[1/1 -1/2 +1/2 -1/3+...+1/n -1/(n+1)]
=n+[1- 1/(n+1)]
=n+ n/(n+1)
=n(n+2)/(n+1)
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