Question

已知正數(shù)數(shù)列{a_n}滿足(n∈N^*),

(1)求a₁,a₂,a₃;

(2)猜測(cè)a_n的表達(dá)式,并證明你的結(jié)論.

Answer 1

解析:=a₁+1(a₁-1)²=0?a₁=1;?

=a₂+1a₂=3;?

=a₃+1a₃=5.?

猜想a_n=2n-1.?

證明:當(dāng)n=1時(shí),顯然成立.?

假設(shè)n=k時(shí)成立,即a_k=2k-1,?

則n=k+1時(shí),=a_k+1+1.                  ①?

∵S_k=()·k,                                            ②

②代入①得(a_k+1-1)²=4k²,?

∴a_k+1=2k+1=2(k+1)-1.?

因此,對(duì)任意自然數(shù)n∈N^*,有a_n=2n-1成立.