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证明:
当n=1时,明显有1/(1×3)=1/(2×1+1)成立,则原式成立。
假设当n=k时,有1/1×3+1/3×5+...+1/(2k-1)(2k+1)=k/2k+1成立,
则当n=k+1时,
1/1×3+1/3×5+...+1/(2k-1)(2k+1)+1/(2(k+1)-1)(2(k+1)+1)
=k/2k+1
+
1/(2k+1)(2k+3)
=k(2k+3)/(2k+1)(2k+3)+1/(2k+1)(2k+3)
=(2k²+3k+1)/(2k+1)(2k+3)
=(2k+1)(k+1)/(2k+1)(2k+3)
=(k+1)/(2(k+1)+1)
综上,由数学归纳法可知,1/1×3+1/3×5+...+1/(2n-1)(2n+1)=n/2n+1(n=1,2,3,……)成立
当n=1时,明显有1/(1×3)=1/(2×1+1)成立,则原式成立。
假设当n=k时,有1/1×3+1/3×5+...+1/(2k-1)(2k+1)=k/2k+1成立,
则当n=k+1时,
1/1×3+1/3×5+...+1/(2k-1)(2k+1)+1/(2(k+1)-1)(2(k+1)+1)
=k/2k+1
+
1/(2k+1)(2k+3)
=k(2k+3)/(2k+1)(2k+3)+1/(2k+1)(2k+3)
=(2k²+3k+1)/(2k+1)(2k+3)
=(2k+1)(k+1)/(2k+1)(2k+3)
=(k+1)/(2(k+1)+1)
综上,由数学归纳法可知,1/1×3+1/3×5+...+1/(2n-1)(2n+1)=n/2n+1(n=1,2,3,……)成立
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