大师作文网一道高一数列题
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一道高一数列题
n>=2,[S(n)]^2 = a(n)[S(n)-1/2],[S(2)]^2 = [S(2)-a(1)][S(2)-1/2] = [S(2)-1][S(2)-1/2] = [S(2)]^2 - 3S(2)/2 + 1/2,S(2)=1/3.[S(n+1)]^2 = a(n+1)[S(n+1)-1/2] = [S(n+1)-S(n)][S(n+1)-1/2] =[S(n+1)]^2 -S(n)S(n+1)-S(n+1)/2+S(n)/2,0=2S(n)S(n+1)+S(n+1)-S(n) 若S(2)=0,0=[S(2)]^2 = a(2)[S(2)-1/2] = -a(2)/2,a(2)=0,0 = S(2)=a(1)+a(2)=1+0=1.矛盾.因此,S(2)不等于0.若S(n+1)=0,n>=2,则由0=2S(n)S(n+1)+S(n+1)-S(n)=-S(n)知S(n）= 0,...,S(2)=0.矛盾.因此,n>=2时,S(n)不等于0.0=2S(n)S(n+1)+S(n+1)-S(n),0 = 2 + 1/S(n) - 1/S(n+1),1/S(n+1) - 1/S(n) = 2,{1/S(n+1)}是首项为1/S(2)=1/(1/3)=3,公差为2的等差数列.1/S(n+1)=3+(n-1)2=2n+1,又1/S(1)=1/a(1)=1,2=1/S(2)-1/S(1).{1/S(n)}是首项为1/S(1)=1,公差为2的等差数列.1/S(n)=2n-1.n=1,2,...S(n) = 1/(2n-1).b(n)=S(n)/(2n+1)=1/[(2n-1)(2n+1)]=(1/2)*[1/(2n-1)-1/(2n+1)].T(n)=b(1)+b(2)+...b(n-1)+b(n) =(1/2)[1/1-1/3 + 1/3-1/5 + ...+ 1/(2n-3)-1/(2n-1) + 1/(2n-1)-1/(2n+1)] = (1/2)[1/1 - 1/(2n+1)] = n/(2n+1).若0 < T(n) - (1/4)(m-8) = n/(2n+1) - (m-8)/4,m-8=2,(m-8)/4>=1/2,-(m-8)/4