求教几个数学数列的问题(超级急)

3个回答

  • 1.

    a(n+1)=3an

    a(n+1)=3an=3*3a(n-1)=3^3*a(n-2)=……=3^(n-1)*a2=3^n*a1=3^(n+1)

    an=3^n

    2.

    a(n+1)=an+1/(4n^2-1)

    =an+1/[(2n+1)(2n-1)]

    =an+(1/2)[(2n-1)-1/(2n+1)]

    a(n+1)+1/(n+1/2)=an+1/(n-1/2)

    a(n+1)+1/(n+1-1/2)=an+1/(n-1/2)=a(n-1)+1/(n-1-1/2)=……=a1+1/(1-1/2)=1/2-2=-3/2

    an+1/(n-1/2)=-3/2

    an=-1/(n-1/2)-3/2

    3.

    a(n+1)=2an+5

    a(n+1)+5=2an+10=2(an+5)

    an+5=(a1+5)2^(n-1)=6*2^(n-1)

    an=6*2^(n-1)-5

    4.

    an=2n+1,数列{bn}中,b1=a1,当n大于等于2时,bn=abn-1(下标是bn-1),则b4=,b5=(此题

    b1=a1

    b2=a[b(2-1)]=a(b1)=a(a1)=a(2+1)=a3=7

    b3=a[b(3-1)]=a(b2)=a7=15

    b4=a[b(4-1)]=a(b3)=a15=31

    b5=a[b(5-1)]=a(b4)=a31=63

    5.

    an=2n/(3n+1)

    =2/3-(2/3)/(3n+1)

    =(2/3)[1-1/(3n+1)]

    是递增数列.

    6.

    ①a1=2,a(n+1)=an+(2n-1)

    a(n+1)=an+(2n-1)

    an=a(n-1)+(2n-3)

    a(n-1)=a(n-2)+(2n-5)

    ……

    a3=a2+(2*3-3)

    a2=a1+(2*2-3)

    叠加:

    an-a1=2[n+(n-1)+……+3+2]-3(n-1)

    =(n+2)(n-1)-3(n-1)

    an=(n+2)(n-1)-3(n-1)+2