（08年赤峰二中模拟理）
数列{an}满足a1 = 2, a1 + a2 + a3 = 12, 且an - 2an + 1 + an + 2 = 0 (n Î N*).
(Ⅱ) 令bn = 
+ 2n - 1 × an, 求数列{bn}的前n项和.
（08年赤峰二中模拟理）
数列{an}满足a1 = 2, a1 + a2 + a3 = 12, 且an - 2an + 1 + an + 2 = 0 (n Î N*).
(Ⅱ) 令bn = + 2n - 1 × an, 求数列{bn}的前n项和.
解析：(Ⅰ)
又a1 + a2 + a3 = 12, 所以 a2 = 4,
因为a1 = 2, 所以an = a1 + (n - 1)(a2 a1) = 2n.
(Ⅱ) 由(Ⅰ)得bn =
+ 2n× n,
令cn =, dn = 2n× n, 则
c1 + c2 + ¼ + cn
= 
+
+ ¼ +
= (1 -
) + (-
) + ¼ + (
-
)
= 1 -,
d1 + d2 + ¼ + dn
= 1 × 21 + 2 × 22 + 3 × 23 + ¼ + n × 2n
2d1 + 2d2 + ¼ + 2dn
= 1 × 22 + 2 × 23 + 3 × 24 + ¼ + n × 2n + 1,
∴ d1 + d2 + ¼ + dn
= - 21 - 22 - 23 - ¼ - 2n + n × 2n + 1
= (n - 1) × 2n + 2,
故b1 + b2 + ¼ + bn
= (c1 + c2 + ¼ + cn) + (d1 + d2 + ¼ + dn)
= (n - 1) × 2n +1 -+ 3.