Question

已知各项均为正数的数列{a_n}满足:在n∈N^*且n＞1时,有=1,a₁=.

(1)求数列{a_n}的通项公式;

(2)证明在n≥5时,a_n≤.

〔参考公式: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满足上式,∴,n∈N^*.

(2)要证原式成立,只需证明n(n+1)(2n+1)≤n·2ⁿ⁺¹+2ⁿ-4n-2,即只需证明n(n+1)(2n+1)≤(2ⁿ-2)(2n+1),即只需证明n(n+1)≤2ⁿ-2,即只需证明n²+n+2≤2ⁿ(n≥5),

因为2ⁿ=(1+1)ⁿ=,

又因为n≥5,故2ⁿ≥=n²+n+2.

则.