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(1-4/1)(1-9/1)(1-16/1).......(1-81/1)(1-1...

解:(1-1/4)(1-1/9).....(1-1/100)=(1-1/2)(1+1/2)(1-1/3)(1+1/3)...(1-1/9)(1+1/9)(1-1/10)(1+1/10)=1/2*3/2*2/3*4/3*3/4*...8/9*10/9*9/10*11/10观察有很多项约掉了,最后得:=1/2*11/10=11/20满意谢谢采纳!

1-1/4=1-(1/2)^=(1-1/2)*(1+1/2)以此类推 原式=(1-1/2)(1+1/2)(1-1/3)(1+1/3)...(1-1/2004)(1+1/2004) 所有相减项=(1/2)*(2/3)(3/4)(4/5)....(2003/2004)相邻相消=1/2004 所有相加项=(3/2)(4/3)(5/4)....(2005/2004)相邻相消=2005/2 所...

11/20 (1-1/4)=1-(1/2)2 =(1+1/2)(1-1/2) (1-1/9)=1-(1/3)2=(1+1/3)(1-1/3) (1-1/16)=1-(1/4)2=(1+1/4)(1-1/4)以此类推(1-1/100)=1-(1/10)2=(1+1/10)(1-1/10) 即3/2*1/2*3/4*2/3*5/4*3/4*6/5*4/5*7/6*5/6*8/7*6/7*9/8*7/8*10/9*8/9*11/10*9/10 ...

1-1/n^2=(n-1)(n+1)/n^2 (1-1/4)(1-1/9)(1-1/16)......(1-1/81)(1-1/100)......(1-1/n的平方) =(1-1/2^2)(1-1/3^2)(1-1/4^2)(1-1/5^2)...(1-1/n^2) =1*3/2^2*2*4/3^2*3*5/4^2*4*6/5^2*...(n-1)(n+1)/n^2 =1*(n+1)/(2*n) =(n+1)/(2n)

原式 =(1-1/2)(1+1/2) * (1-1/3)(1+1/3) * ... * (1-1/10)(1+1/10) =1/2 * 2/3 * 3/4 * ... * 9/10 * 3/2 * 4/3 * 5/4 * ... * 11/10 整理,抵消 =1/2 * 11/10 =11/20

(n+1)/2n

.................

=(1-1/2)(1+1/2)(1-1/3)(1+1/3)(1-1/4)(1+1/4)...(1-1/100)(1+1/100) =(1/2)(3/2)(2/3)(4/3)(3/4)(5/4)...(99/100)(101/100) =(1/2)(101/100) =101/200

a1=(1-1/4)=3/4 a2=(1-1/4)(1-1/9) =3/4*8/9 =1/3=4/12 a3=a2*(1-1/16)=1/3*15/16=5/16 a4=a3*(1-1/25)=5/16*24/25=3/10=6/20 所以通项从第二项开始猜测为分母=n+2 分子=4(n+1) 所以an=(n+2)/[4(n+1)],n>=2 a1=3/4 对于n=2显然成立 假设对于n=k...

(1-1/4)(1-1/9)(1-16)(1-1/25)...(1-1/10000) =(1-1/2)(1+1/2)(1-1/3)(1+1/3)...(1-1/100)(1+1/100) =(1/2)(3/2)(2/3)(4/3)(3/4)...(100/99)(99/100)(101/100) =(1/2)(101/100) =101/200

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