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求人大第五版&&高级会计学&&课件ppt，下学期要用
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假如爱有天意！
呵呵，已经在人大出版社下载了。
楼上的&&人大出版社& &下载需要什么吗/
楼上坐板凳的，通过教师认证，即可在人大出版社下载
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论坛法律顾问：王进律师已知数列{an}满足a1=1,a(n+1)=an+√((an)^2+1),令an=tanθn(0&θn&π/2),求证（1）数列{θn-π/2}是等比数列（2）a1+a2+…+an&(n-1)π/2
证明：(1)a(n+1)=an+√((an)^2+1)a(n+1)=tan(θ(n+1))an+√((an)^2+1)=tan(θn)+√(tan^2(θn)+1)=tan(θn)+1/(cos(θn))=(sin(θn)+1)/(cos(θn))=(sin(θn)+sin^2(θn/2)+cos^2(θn/2))/(cos(θn))=(2*sin(θn/2)*cos(θn/2)+sin^2(θn/2)+cos^2(θn/2))/(cos^2(θn/2)-sin^2(θn/2))=(sin(θn/2)+cos(θn/2))^2/((sin(θn/2)+cos(θn/2))(cos(θn/2)-sin(θn/2)))=(sin(θn/2)+cos(θn/2))/(cos(θn/2)-sin(θn/2)))=(tan(θn/2)+1)/(1-tan(θn/2))=tan(θn/2+π/4)即θ(n+1)=θn/2+π/4θ(n+1)-π/2=(1/2)*(θn-π/2)故{θn-π/2}是等比数列得证(2)a1=tan(θ1)=10(n-1)*π/2 得证
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