1.
tan(2α)=2tanα/(1-tan²α)=2(√2-1)/[1-(√2-1)²]=2(√2-1)/(1-3+2√2)=2(√2-1)/[2(√2-1)]=1
α为锐角,2α0,假设当n=k(k∈N+)时,ak>0,则当n=k+1时,
a(k+1)=ak²+ak>0,k为任意正整数,因此对于任意正整数n,an恒>0
a(n+1)-an=an²+an-an=an²>0
a(n+1)>an
3.
a2=a1²+a1=(1/2)²+1/2=3/4
1/(1+a1)+1/(1+a2)=1/(1+1/2)+1/(1+3/4)=2/3 +4/7=26/21>1
an>0 1+an>1>0
1/(1+a1)+1/(1+a2)+...+1/(1+an)≥26/21+0+...+0=26/21>1
当且仅当n=2时1/(1+a1)+1/(1+a2)+...+1/(1+an)≥26/21取等号
1/(1+a1)+1/(1+a2)+...+1/(1+an)>1
1/a(n+1)=1/(an²+an)=1/[an(an +1)]=1/an -1/(an +1)
1/(an +1)=1/an -1/a(n+1)
1/(1+a1)+1/(1+a2)+...+1/(1+an)
=1/a1-1/a2+1/a2-1/a3+...+1/an -1/a(n+1)
=1/a1 -1/a(n+1)
=1/(1/2)-1/a(n+1)
=2- 1/a(n+1)
an>0 a(n+1)>0 1/a(n+1)>0
2- 1/a(n+1)