题目

已知等差数列a1=3/2,公差为1,设bn=a*2的n次方+b*an-75(a,b属于自然数),且数列{bn}的前项和Tn的最小值是T6求a,b的值.

提问时间：2021-12-23

答案

an=3/2+(n-1)
=n+1/2
bn=a*2^n+b*an-75
=a*2^n+b(n+1/2)-75
=a*2^n+b*n+b/2-75
Tn=[a*2+b*1+b/2-75]+[a*2^2+b*2+b/2-75]+……+[a*2^(n-2)+b*(n-2)+b/2-75]+[a*2^(n-1)+b*(n-1)+b/2-75]+[a*2^n+b*n+b/2-75]
=a[2+2^2+……+2^(n-2)+2^(n-1)+2^n]+b[1+2+3+……+(n-2)+(n-1)+n]+nb/2-75n
=2a(2^n-1)+n(n+1)b/2+nb/2-75n
=a*2^(n+1)+(b/2)n^2+(b-75)n-2a
Tn=a*2^(n+1)+(b/2)n^2+(b-75)n-2a
T6=a*2^7+18b+6b-450-2a
=126a+24b-450
a*2^(n+1)+(b/2)n^2+(b-75)n-2a≥126a+24b-450