Question

已知各項均為正數(shù)的數(shù)列{a_n}滿足:在n∈N^*且n＞1時,有=1,a₁=.

(1)求數(shù)列{a_n}的通項公式;

(2)證明在n≥5時,a_n≤.

〔參考公式:1²+2²+3²+…+n²=n(n+1)(2n+1)(n∈N^*)〕

Answer 1

解：(1)由已知有(n∈N^*且n＞1),又a₁²=6,

∴a_n²=(a_n²-a_n-1²)+(a_n-1²-a_n-2²)+…+(a₂²-a₁²)+a₁²=6［n²+(n-1)²+…+2²+1²］=n(n+1)(2n+1)(n∈N^*,n＞1).

又a₁²=6滿足上式,∴,n∈N^*.

(2)要證原式成立,只需證明n(n+1)(2n+1)≤n·2ⁿ⁺¹+2ⁿ-4n-2,即只需證明n(n+1)(2n+1)≤(2ⁿ-2)(2n+1),即只需證明n(n+1)≤2ⁿ-2,即只需證明n²+n+2≤2ⁿ(n≥5),

因為2ⁿ=(1+1)ⁿ=,

又因為n≥5,故2ⁿ≥=n²+n+2.

則.