Question

己知数列｛an｝是首项a1=1/2,公比q=1/2的等比数列,设bn+2=3log1/2an,数列｛Cn｝满足Cn=an

Answer 1

(1)an=(1/2)^n 3log1/2an=3 bn=3n-2

(2)sn=(1/2)*1+(1/2)^2*4+(1/2)^3*7+...+(1/2)^n*(3n-2)

2sn=1+(1/2)^1*4+(1/2)^2*7+...+(1/2)^(n-1)*(3n-2)

(2-1)sn=1-3[(1/2)^1+(1/2)^2+...+(1/2)^(n-1)]-(1/2)^n(3n-2)

=(8-3n)(1/2)^n-2

(3)(1/2)^n*(3n-2)>=(1/2)^(n+1)*(3n+1)且(1/2)^n*(3n-2)>=(1/2)^(n-1)*(3n-5)

得(Cn)max=C2=1

只需使2m^2+3m-4>=1

解得m=1