简算:[(1*2*4+2*4*8+L+n.2n.4n)/(1*3*9+2*6*18+L=n.3n.9n)]^2
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/05/11 03:51:32
简算:[(1*2*4+2*4*8+L+n.2n.4n)/(1*3*9+2*6*18+L=n.3n.9n)]^2
简算:
1.[(1*2*4+2*4*8+L+n.2n.4n)/(1*3*9+2*6*18+L=n.3n.9n)]^2
2.(1/11*13*15)+(1/13*15*17)+L+(1/29*31*33)
3.(1+1/1*3)(1+1/2*4)(1+1/3*5)L(1+1/1998*2000)(1+1/1999*2001)
4.(4/1*3)-(8/3*5)+(12/5*7)-(16/7*9)+L-(40/19*21)
5.1+2+2^2+2^3+2^4+L+2^2010
6.1+(1/1+2)+(1/1+2+3)+(1/1+2+3+4)+L+(1/1+2+3+4+...+100)
注解:“L”等于省略号,(X/Y)=Y分之X
简算:
1.[(1*2*4+2*4*8+L+n.2n.4n)/(1*3*9+2*6*18+L=n.3n.9n)]^2
2.(1/11*13*15)+(1/13*15*17)+L+(1/29*31*33)
3.(1+1/1*3)(1+1/2*4)(1+1/3*5)L(1+1/1998*2000)(1+1/1999*2001)
4.(4/1*3)-(8/3*5)+(12/5*7)-(16/7*9)+L-(40/19*21)
5.1+2+2^2+2^3+2^4+L+2^2010
6.1+(1/1+2)+(1/1+2+3)+(1/1+2+3+4)+L+(1/1+2+3+4+...+100)
注解:“L”等于省略号,(X/Y)=Y分之X
算你狠,这么多题目!
再问: 。。。。
再答: 算你狠,这么多题目!! 还没有加分!!!
再问: 差点忘了。谢谢提醒啊。
再答: 简算: 1.[(1*2*4+2*4*8+L+n.2n.4n)/(1*3*9+2*6*18+L=n.3n.9n)]^2 ={[1*2*4*(1+2+……+n)]/[1*3*9*(1+2+……+n)]}^2 =[(1*2*4)/(1*3*9)]^2 =(8/9)^2 =64/81 2.(1/11*13*15)+(1/13*15*17)+L+(1/29*31*33) =(1/11*13-1/13*15+1/13*15-1/15*17+……+1/29*31-1/31*33)÷2 =(1/11*13-1/31*33)÷2 =80/13299÷2 =40/13299 3.(1+1/1*3)(1+1/2*4)(1+1/3*5)L(1+1/1998*2000)(1+1/1999*2001) 1+1/n(n+2)=(n+1)^2/n(n+2) 即(2*2/1*3)(3*3/2*4)*(4*4/3*5)...............(1999^2/1998*2000)*(2000^2)/1999*2001 错位消项(分母分为2部分,对着分子消) 1*2*3*.......1999 3*4*5*.......*2000 原式=2*2000/2001=4000/2001 4.(4/1*3)-(8/3*5)+(12/5*7)-(16/7*9)+L-(40/19*21) =1+1/3-1/3-1/5+1/5+1/7-1/7-1/9+……-1/19-1/21 =1-1/21 =20/21 5.1+2+2^2+2^3+2^4+L+2^2010 =(2^2011-1)/(2-1) =2^2011-1 6.1+(1/1+2)+(1/1+2+3)+(1/1+2+3+4)+L+(1/1+2+3+4+...+100) =1+2/(2x3)+2/(3x4)+2/(4x5)+……+2/(100x101) =1+2x(1/2-1/3+1/3-1/4+1/4-1/5+……+1/100-1/101) =1+2x(1/2-1/101) =1+1-2/101 =200/101
再问: 。。。。
再答: 算你狠,这么多题目!! 还没有加分!!!
再问: 差点忘了。谢谢提醒啊。
再答: 简算: 1.[(1*2*4+2*4*8+L+n.2n.4n)/(1*3*9+2*6*18+L=n.3n.9n)]^2 ={[1*2*4*(1+2+……+n)]/[1*3*9*(1+2+……+n)]}^2 =[(1*2*4)/(1*3*9)]^2 =(8/9)^2 =64/81 2.(1/11*13*15)+(1/13*15*17)+L+(1/29*31*33) =(1/11*13-1/13*15+1/13*15-1/15*17+……+1/29*31-1/31*33)÷2 =(1/11*13-1/31*33)÷2 =80/13299÷2 =40/13299 3.(1+1/1*3)(1+1/2*4)(1+1/3*5)L(1+1/1998*2000)(1+1/1999*2001) 1+1/n(n+2)=(n+1)^2/n(n+2) 即(2*2/1*3)(3*3/2*4)*(4*4/3*5)...............(1999^2/1998*2000)*(2000^2)/1999*2001 错位消项(分母分为2部分,对着分子消) 1*2*3*.......1999 3*4*5*.......*2000 原式=2*2000/2001=4000/2001 4.(4/1*3)-(8/3*5)+(12/5*7)-(16/7*9)+L-(40/19*21) =1+1/3-1/3-1/5+1/5+1/7-1/7-1/9+……-1/19-1/21 =1-1/21 =20/21 5.1+2+2^2+2^3+2^4+L+2^2010 =(2^2011-1)/(2-1) =2^2011-1 6.1+(1/1+2)+(1/1+2+3)+(1/1+2+3+4)+L+(1/1+2+3+4+...+100) =1+2/(2x3)+2/(3x4)+2/(4x5)+……+2/(100x101) =1+2x(1/2-1/3+1/3-1/4+1/4-1/5+……+1/100-1/101) =1+2x(1/2-1/101) =1+1-2/101 =200/101
简算:[(1*2*4+2*4*8+L+n.2n.4n)/(1*3*9+2*6*18+L=n.3n.9n)]^2
简算:1.[(1*2*4+2*4*8+L+n.2n.4n)/(1*3*9+2*6*18+L=n.3n.9n)]^22.(
1\n(n+3)+1\(n+3)(n+6)+1\(n+6)(n+9)=1\2 n+18 n为正整数,求n的值
证明不等式:(1/n)^n+(2/n)^n+(3/n)^n+.+(n/n)^n
lim 9^n+4^n+2/5^n-3^2n-1 n趋于无穷大时
Sn=n(n+2)(n+4)的分项等于1/6[n(n+2)(n+4)(n+5)-(n-1)n(n+2)(n+4)]吗?
化简:1/(n+1)(n+2)+1/(n+2)(n+3)+1/(n+3)(n+4)
(n+1)(n+2)/1 +(n+2)(n+3)/1 +(n+3)(n+4)/1
证明:1+2C(n,1)+4C(n,2)+...+2^nC(n,n)=3^n .(n∈N+)
计算【(1*2*4+2*4*8+...+n*2n*4n)/(1*3*9+2*6*18+...+n*3n*9n)】的2次方
求[(1*2*4+2*4*8+…+n*2n*4n)/(1*3*9+2*6*18+...+n*3n*9n)]^2
(1*2*4+2*4*8+……+n*2n*4n/1*3*9+2*6*18+……+n*3n*9n)^2