Question

已知数列{an}满足an+1+an=4n-3

1 若数列an是等差数列，求a1的值
2 当a1=1时，求数列an的前n项和Sn

Answer 1

解：1、
若数列an是等差数列
可设a(n+1)=a1+nd
an=a1+(n-1)d
a(n+1)+an=2a1+(2n-1)d=4n-3
整理，得
2dn+2a1-d=4n-3
2d=4
2a1-d=-3
解得
d=2 a1=-1/2
a1=-1/2
2、
因为a(n+1) + an = 4n -3
即a(n+1) - 2(n+1) + 5/2 = - ( an - 2n + 5/2 )
令 bn = an - 2n + 5/2
于是b1 = a1 - 2 + 5/2 = 5/2
于是b(n+1) = - bn
∴bn = 5/2 *(-1)^(n+1）
an = bn +2n - 5/2 = 5/2 [ (-1)^(n+1) - 1 ] + 2n
当n为奇数时，an = 2n
当n为偶数时，an = 2n - 5
当n为奇数时，Sn = (2+4+6+……+2n) - (n-1)/2 *5 = n² - 3/2 n + 5/2
当n为偶数时，Sn = (2+4+6+……+2n) - n/2 *5 = n² - 3/2 n
可合并为 Sn = n² - 3/2 n + 5/4 * [ (-1)^(n+1) + 1]