Question

高2数学题

Answer 1

1.Sn/Tn=(2a1-d1+nd1)/(2b1-d2+nd2)=(5n+3)/(2n+7)nbsp;所以d1=5Knbsp;a1=4Knbsp;d2=2Knbsp;b1=4.5Knbsp;K为不等于0的系数nbsp;a5=24Knbsp;b5=12.5Knbsp;a5/b5=24/12.5=48/25nbsp;2.设an=2q^(n-1)nbsp;{an+1}也是等比数列，故(an+1)^=[a(n-1)+1][a(n+1)+1]nbsp;[2q^(n-1)+1]=[2q^(n-2)+1][2q^n+1]nbsp;4q^(2n-2)+4q^(n-1)+1nbsp;=nbsp;4q^(2n-2)+2q^n+2q^(n-2)+1nbsp;2q^(n-1)nbsp;=nbsp;q^nnbsp;+nbsp;q^(n-2)..........q≠0nbsp;q^-2q+1=0=(q-1)^nbsp;从而q=1故{An}是常数列，an=2nbsp;故Sn=2nnbsp;3.由已知：ac=b^2;a+b=2x;b+c=2y;nbsp;得到：(a+b)/x+(b+c)/y=4;nbsp;即：a/x+c/y+b/x+b/y=4nbsp;……（1）nbsp;又有ac=(2x-b)(2y-b)=b^2;nbsp;化简：2xy=bx+by;nbsp;即：b/x+b/y=2nbsp;……（2）nbsp;把2带入1中，得a/x+c/y=2.nbsp;4.因为a(n+1)=2an+3nbsp;；所以a(n+1)+3=2(an+3）nbsp;[a(n+1)+3]/(an+3）=2nbsp;{an+3}是以a1+3=4位首项，2为公比的等比数列nbsp;an+3=4*2^(n-1)nbsp;故通项an=4*2^(n-1)-3nbsp;5.nbsp;∵a(n+1)=an+n+1nbsp;∴a(n+1)-an=n+1nbsp;an-a(n-1)=(n-1)+1=n...nbsp;(1)nbsp;a(n-1)-a(n-2)=(n-2)+1=n-1nbsp;...nbsp;(2)nbsp;...nbsp;...nbsp;a3-a2=3nbsp;a2-a1=2...nbsp;(n)nbsp;(1)+(2)+......+(n)得:nbsp;an-a1=2+3+...+n=n(n+1)/2-1nbsp;故：a100=100(100+1)/2-1+a1=5051nbsp;6.Sn=n^2,故S(n-1)=(n-1)^2nbsp;故：an=Sn-S(n-1)=n^2-(n-1)^2=2n-1.a1=1^2=1.从而a2=3,a3=5,a4=7即三边之比为3：5：7，设为a,b,c，由余弦定理得cosC=(c^2-b^2-c^2)/2bc=1/2从而C=arccos(1/2)=60°。终于好了！