Question

已知数列{an}的通项公式是an=(3n-1)(4/5)^n,求数列{an}的最大项

Answer 1

a(n+1)/a(n)=[3n+2](4/5)/[3n-1],
a(n)>0.
若a(n+1)/a(n)>1,则 4(3n+2)/[5(3n-1)]>1, 4(3n+2)>5(3n-1), 13>3n, 13/3>n, 4>=n.
若a(n+1)/a(n)<1,则 4(3n+2)/[5(3n-1)]<1, 4(3n+2)<5(3n-1), 13<3n, 13/3<n, 4<n.
n<=4时,a(n+1)/a(n)>1, a(n+1)>a(n), a(n)<=a(5)=(3*5-1)*(4/5)^5=14*(4/5)^5.
n>4时, a(n+1)/a(n)<1, a(n+1)<a(n), a(n)<=a(5).
综合,有,
a(n)<=a(5).
{a(n)}的最大项为a(5)=14*(4/5)^5