求和1/1·3+1/3·5+1/5·7+...+1/(2n+1)(2n-1)
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求和1/1·3+1/3·5+1/5·7+...+1/(2n+1)(2n-1)
1/(2n+1)(2n-1)
=[(2n+1)-(2n-1)]/【2×(2n+1)(2n-1)】
=1/2×[(2n+1)-(2n-1)]/【(2n+1)(2n-1)】
=1/2×[1/(2n-1)-1/(2n+1)
所以1/1*3=1/2×(1-1/3)
1/3*5=1/2×(1/3-1/5)
.
所以原式=1/2×(1-1/3)+1/2*(1/3-1/5)+.+1/2×[1/(2n-1)-1/(2n+1)
=1/2*(1-1/3+1/3-1/5+1/5-1/7+.+1/(2n-1)-1/(2n+1))
=1/2*(1-1/(2n+1))
=1/2*2n/(2n+1)
=n/2n+1
=[(2n+1)-(2n-1)]/【2×(2n+1)(2n-1)】
=1/2×[(2n+1)-(2n-1)]/【(2n+1)(2n-1)】
=1/2×[1/(2n-1)-1/(2n+1)
所以1/1*3=1/2×(1-1/3)
1/3*5=1/2×(1/3-1/5)
.
所以原式=1/2×(1-1/3)+1/2*(1/3-1/5)+.+1/2×[1/(2n-1)-1/(2n+1)
=1/2*(1-1/3+1/3-1/5+1/5-1/7+.+1/(2n-1)-1/(2n+1))
=1/2*(1-1/(2n+1))
=1/2*2n/(2n+1)
=n/2n+1
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