高一数学数列问题~~~~~请各位高手帮帮忙数列{aˇn}中,aˇ1=1,当n>=2时,其前n项和Sˇn满足S^2=aˇn(Sˇn-1/2) (1)求Sˇn得aˇn (2)设bˇn=Sˇn/(2n+1),数列{bˇn}的前n项和为Tˇn,求Tˇn
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![高一数学数列问题~~~~~请各位高手帮帮忙数列{aˇn}中,aˇ1=1,当n>=2时,其前n项和Sˇn满足S^2=aˇn(Sˇn-1/2) (1)求Sˇn得aˇn (2)设bˇn=Sˇn/(2n+1),数列{bˇn}的前n项和为Tˇn,求Tˇn](/uploads/image/z/693772-52-2.jpg?t=%E9%AB%98%E4%B8%80%E6%95%B0%E5%AD%A6%E6%95%B0%E5%88%97%E9%97%AE%E9%A2%98%7E%7E%7E%7E%7E%E8%AF%B7%E5%90%84%E4%BD%8D%E9%AB%98%E6%89%8B%E5%B8%AE%E5%B8%AE%E5%BF%99%E6%95%B0%E5%88%97%7Ba%CB%87n%7D%E4%B8%AD%2Ca%CB%871%3D1%2C%E5%BD%93n%3E%3D2%E6%97%B6%2C%E5%85%B6%E5%89%8Dn%E9%A1%B9%E5%92%8CS%CB%87n%E6%BB%A1%E8%B6%B3S%5E2%3Da%CB%87n%28S%CB%87n-1%2F2%29+%281%29%E6%B1%82S%CB%87n%E5%BE%97a%CB%87n++++++++++++++++++++++++++++%282%29%E8%AE%BEb%CB%87n%3DS%CB%87n%2F%282n%2B1%29%2C%E6%95%B0%E5%88%97%7Bb%CB%87n%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BAT%CB%87n%2C%E6%B1%82T%CB%87n)
高一数学数列问题~~~~~请各位高手帮帮忙数列{aˇn}中,aˇ1=1,当n>=2时,其前n项和Sˇn满足S^2=aˇn(Sˇn-1/2) (1)求Sˇn得aˇn (2)设bˇn=Sˇn/(2n+1),数列{bˇn}的前n项和为Tˇn,求Tˇn
高一数学数列问题~~~~~请各位高手帮帮忙
数列{aˇn}中,aˇ1=1,当n>=2时,其前n项和Sˇn满足S^2=aˇn(Sˇn-1/2)
(1)求Sˇn得aˇn (2)设bˇn=Sˇn/(2n+1),数列{bˇn}的前n项和为Tˇn,求Tˇn
高一数学数列问题~~~~~请各位高手帮帮忙数列{aˇn}中,aˇ1=1,当n>=2时,其前n项和Sˇn满足S^2=aˇn(Sˇn-1/2) (1)求Sˇn得aˇn (2)设bˇn=Sˇn/(2n+1),数列{bˇn}的前n项和为Tˇn,求Tˇn
(1)
n=1:s(1)=a(1)
n>=2:a(n)=s(n)-s(n-1)
带入S^2=a(n)(S-1/2)中,得:
S^2=[S(n)-S(n-1)]*[S(n)-1/2]
整理得到:
S(n)=S(n-1)/[1+2S(n-1)]
则S(1)=1
S(2)=1/3
S(3)=1/5
有数学归纳法可证明,S(n)=1/(2n-1)
n=1,a(1)=1
n>=2时
S(n-1)=1/(2n-3)
a(n)=s(n)-s(n-1)=1/(2n-1)-1/(2n-3)=2/[(2n-1)(2n-3)]
(2)
b(n)=S(n)/(2n+1)=1/(2n-1)(2n+1)=1/2*[1/(2n-1)-1/(2n+1)]
则T(n)=b(1)+b(2)+...+b(n)=1/2*[1-1/3+1/3-1/5+.-1/(2n+1)]=1/2*[1-1/(2n+1)]=n/2n+1