大师作文网求和1/1*5 + 1/3*7 +1/5*9 +1/7*11+.+1/(2n-1)(2n+3)
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求和1/1*5 + 1/3*7 +1/5*9 +1/7*11+.+1/(2n-1)(2n+3)
如题 顺便说明一下原理
=1/3-(n+1)/(2n+1)(2n+3)
=n(4n+5)/3(2n+1)(2n+3) 请问这两步是怎么回事？
如题 顺便说明一下原理
=1/3-(n+1)/(2n+1)(2n+3)
=n(4n+5)/3(2n+1)(2n+3) 请问这两步是怎么回事？
1/(2n-1)(2n+3)=1/4*[1/(2n-1)-1/(2n+3)]
所以
1/1*5+1/3*7+1/5*9+1/7*11+.1/(2n-1)*(2n+3)
=1/4*(1-1/5)+1/4*(1/3-1/7)+1/4*(1/5-1/9)+1/4*(1/7-1/11)+1/4*[1/(2n-1)-1/(2n+3)]
=1/4*[(1-1/5)+(1/3-1/7)+(1/5-1/9)+(1/7-1/11)+.+1/(2n-1)-1/(2n+3)]
=1/4*[1+1/3+1/(2n+1)-1/(2n+3)]
=1/3-(n+1)/(2n+1)(2n+3)
=n(4n+5)/3(2n+1)(2n+3)
所以
1/1*5+1/3*7+1/5*9+1/7*11+.1/(2n-1)*(2n+3)
=1/4*(1-1/5)+1/4*(1/3-1/7)+1/4*(1/5-1/9)+1/4*(1/7-1/11)+1/4*[1/(2n-1)-1/(2n+3)]
=1/4*[(1-1/5)+(1/3-1/7)+(1/5-1/9)+(1/7-1/11)+.+1/(2n-1)-1/(2n+3)]
=1/4*[1+1/3+1/(2n+1)-1/(2n+3)]
=1/3-(n+1)/(2n+1)(2n+3)
=n(4n+5)/3(2n+1)(2n+3)
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