(1)2Sn=a(n+1)-2^(n+1)+1令n=1,2联立(a2+5)*2=a1+a3得a1=1
(2)2an=2sn-2sn-1=a(n+1)-an-2^n
即a(n+1)=3an+2^n
所以a(n+1)+2^(n+1)=3*(an+2^n)
an+2^n=(a1+2^1)*3^(n-1)=3^n
an=3^n-2^n
(3)证明只要证1/a1+1/a2+...1/an
(1)2Sn=a(n+1)-2^(n+1)+1令n=1,2联立(a2+5)*2=a1+a3得a1=1
(2)2an=2sn-2sn-1=a(n+1)-an-2^n
即a(n+1)=3an+2^n
所以a(n+1)+2^(n+1)=3*(an+2^n)
an+2^n=(a1+2^1)*3^(n-1)=3^n
an=3^n-2^n
(3)证明只要证1/a1+1/a2+...1/an