等差数列{an}是递增数列,前n项和为Sn,且a1,a3,a9成等比数列,S5=(a5)^2

1个回答

  • (1)由等差数列通项公式和求和公式:

    an=a1+(n-1)*d

    Sn=n*a1+1/2 [n*(n-1)]*d

    a3^2=a1*a9

    S5=(a5)^2

    (a1+2d)^2=a1(a1+8d)

    5a1+10d=(a1+4d)^2

    解得

    a1= d=3/5

    或 a1=d=0

    又因为an为递增数列,d不为0

    所以 an的通项公式为

    an=3/5+3/5*(n-1)=3n/5

    (2)题目写的不太清楚!

    因为

    bn=(n^2+n+1)/[an*a(n+1)] =(n^2+n+1)/[(9/25)n(n+1)]

    = 25/9* (n^2+n+1)/(n^2+n)

    = 25/9 + 25/9 * 1/n(n+1)

    = 25/9 + 25/9[1/n - 1/(n+1)]

    则数列{bn}的前n项和为

    Sn = b1 +b2+ b3 +.+bn

    = 25n/9 + 25/9[1 - 1/(n+1)]

    所以数列bn的前99项的和为

    S99 = 25*11 + 25/9 * 99/100 = 1111/4