Question

已知数列an的前n项和Sn=-n²╱2+kn,且Sn的最大值为8,（1）确定常数k,并求an,（2）求数列（9

Answer 1

(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)