已知数列{an}的前n项和Sn=-1/2n^2+kn,k∈N*,且Sn的最大值为8.1)确定常数k,

1个回答

  • (1)

    Sn=-n²/2 +kn

    =(-1/2)(n²-2kn+k²)+k²/2

    =(-1/2)(n-k)²+k²/2

    当n=k时,Sn有最大值(Sn)max=k²/2=8

    k²=16

    k=-4(k为自然数,舍去)或k=4

    k=4

    (2)

    Sn=-n²/2 +4n

    n=1时,a1=S1=-1/2 +4=7/2

    n≥2时,

    Sn=-n²/2 +4n S(n-1)=-(n-1)²/2 +4(n-1)

    an=Sn-S(n-1)=-n²/2 +4n +(n-1)²/2 -4(n-1)

    = -n +9/2

    =(9-2n)/2

    n=1时,a1=(9-2)/2=7/2,同样满足.

    数列{an}的通项公式为an=(9-2n)/2

    (9-2an)/2ⁿ=[9-2(9-2n)/2]/2ⁿ

    =n/ 2^(n-1)

    Tn=1/2^0 +2/2^1+3/2^2+...+n/2^(n-1)

    Tn/2=1/2^1+2/2^2+...+(n-1)/2^(n-1)+n/2ⁿ

    Tn-Tn/2=Tn/2=1/2^0+1/2^1+1/2^2+...+1/2^(n-1) -n/2ⁿ

    =[1-(1/2)ⁿ]/(1-1/2) -n/2ⁿ

    =2-2/2ⁿ-n/2ⁿ

    =2-(n+2)/2ⁿ

    Tn=4- (n+2)/2^(n-1)