Question

（08年赤峰二中模拟理）

数列{a_n}满足a₁ = 2, a₁ + a₂ + a₃ = 12, 且a_n - 2a_{n + 1} + a_{n + 2} = 0 (n Î N*).

       (Ⅰ) 求数列{a_n}的通项公式;

   (Ⅱ) 令b_n = + 2^{n - 1} × a_n, 求数列{b_n}的前n项和.

Answer 1

解析：(Ⅰ) 因为a_n - 2a_{n + 1} + a_{n + 2} = 0 (n Î N*), 所以数列{a_n}为等差数列,

又a₁ + a₂ + a₃ = 12, 所以 a₂ = 4,

因为a₁ = 2, 所以a_n = a₁ + (n - 1)(a₂ a₁) = 2n.

(Ⅱ) 由(Ⅰ)得b_n =+ 2ⁿ× n,

令c_n =, d_n = 2ⁿ× n,  则

 c₁ + c₂ + ¼ + c_n

= ++ ¼ +

= (1 -) + (-) + ¼ + (-)

= 1 -,

d₁ + d₂ + ¼ + d_n

= 1 × 2¹ + 2 × 2² + 3 × 2³ + ¼ + n × 2ⁿ

2d₁ + 2d₂ + ¼ + 2d_n

= 1 × 2² + 2 × 2³ + 3 × 2⁴ + ¼ + n × 2^{n + 1},

∴ d₁ + d₂ + ¼ + d_n

= - 2¹ - 2² - 2³ - ¼ - 2ⁿ + n × 2^{n + 1}

= (n - 1) × 2ⁿ + 2,

故b₁ + b₂ + ¼ + b_n

= (c₁ + c₂ + ¼ + c_n) + (d₁ + d₂ + ¼ + d_n)

= (n - 1) × 2^{n +1}  -+ 3.