Question

数列{an} a1=0,a2=2,an+2=(1+cos²nл/2)an+4sin²nл/2,n=1,2,3…… 求a3

Answer 1

a(n+2)=(1+cos²nл/2)an+4sin²nл/2

=[1+(1+cosnπ)/2]an+2(1-cosnπ)

①当n∈2N时,a(n+2)=2an

②当n∈2N+1时,a(n+2)=an+4

a3=a1+4=4

a4=2*a2=4

a5=a3+4=8

a6=2*a4+4=12

Sk=a1+a3+a5+...+a(2k+1)

=(k+1)a1+k(k+1)*4/2

=2k²+2k

Tk=a2+a4+a6+...+a(2k)

=a2*(2^k-1)(2-1)

=2^(k+1)-2

Wk=2^Sk/(2+Tk)=2^(2k²+2k)/2^(k+1)>1

2^(2k²+2k)>2^(k+1)

2k²+2k>k+1

2k²+k-1>0

(2k-1)(k+1)>0

k1/2

∴k≥1(k∈N)