Sn=1/1×3+1/3×5+1/5×7+…+1/(2n-1)(2n+1)求和
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Sn=1/1×3+1/3×5+1/5×7+…+1/(2n-1)(2n+1)求和
Sn=1/1×3+1/3×5+1/5×7+…+1/(2n-1)(2n+1)
求
和
Sn=1/1×3+1/3×5+1/5×7+…+1/(2n-1)(2n+1)求和
Sn=1/1×3+1/3×5+1/5×7+…+1/(2n-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)
Sn=(1/2)[1-(1/3)+(1/3-(1/5)+……
+(1/(2n-1))-(1/((2n+2))]=1/2-1/(4n+2)
Sn=1/1×3+1/3×5+1/5×7+…+1/(2n-1)(2n+1)=(1/2)[(1-1/3)+(1/3-1/5)+……+1/(2n-1))-1/((2n+1)]=1/2-1/(4n+2)=(2n+1-1)/2(2n+1)
=n/(2n+1)
(1/3*5)=(1/3-1/5)/2
1/(2n+1)(2n+3)=(1/(2n+1)-1/(2n+3))/2
所以
Sn=(1/3+1/3*5)+(1/5*7)+(1/7*9)+...+[1/(2n+1)(2n+3)]
=0.5*[1/3-1/5+1/5-1/7+...+1/(2n+1)-1/(2n+3)]
=0.5*[1/3-1/(2n+3)] + 1/3
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