Question

設(shè)數(shù)列{a_n}滿足:a_n+1 =a_n²-na_n+1,n=1,2,3,…,

(1)當(dāng)a₁=2時(shí),求a₂,a₃,a₄,并由此猜想出a_n的一個(gè)通項(xiàng)公式;

(2)當(dāng)a₁≥3時(shí),證明對(duì)所有的n≥1,有

(ⅰ)a_n≥n+2;

(ⅱ) +…+.

Answer 1

(1)解析:由a₁=2,a₂=a₁²-a₁+1=3,?

∵a₂=3,a₃=a₂²-2a₂+1=4,?

由a₃=4,∴a₄=a₃²-3a₃+1=5.?

由此猜想a_n的一個(gè)通項(xiàng)公式:a_n=n+1(x≥1).?

(2)證明:(ⅰ)用數(shù)學(xué)歸納法證明:?

①當(dāng)n=1時(shí),a₁≥3=1+2,不等式成立.?

②假設(shè)當(dāng)n=k時(shí)不等式成立,即?

a_k≥k+2,那么?

a_k+1 =a_k(a_k-k)+1≥(k+2)(k+2-k)+1≥k+3,?

即當(dāng)n=k+1時(shí),a_k+1≥(k+1)+2.?

根據(jù)①和②,對(duì)于所有n≥1,a_n≥n+2.??

(ⅱ)由a_n+1=a_n(a_n-n)+1及(ⅰ),對(duì)k≥2,a_k=a_k-1(a_k-1-k+1)+1?

≥a_k-1(k-1+2-k+1)+1?

=2a_k-1+1,……?

∴a_k≥2_k-1a₁+2_k-2+…+2+1=2_k-1(a₁+1)-1.?

于是.?

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