Question

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T

Answer 1

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