高中数学数列题 急求答案已知数列an的首项a1=a,其前n项和为Sn,且满足Sn+Sn-1=3n^2(n≥2).若对任意的n∈N* ,an＜an+1恒成立,则a的取值范围是
2019-04-14
高中数学数列题 急求答案
已知数列an的首项a1=a,其前n项和为Sn,且满足Sn+Sn-1=3n^2(n≥2).若对任意的n∈N* ,an＜an+1恒成立,则a的取值范围是
优质解答
Sn + Sn-1 = 3n^2
Sn+1 + Sn = 3(n+1)^2
Sn+1 - Sn-1 = 3(n+1)^2 - 3n^2 = 6n+3
即 an+1 + an = 6n+3
an+2 + an+1 = 6(n+1)+3=6n+9
an+2 - an = 6
所以,任何情况下an < an+2,且 an+2 - an = 6
所以只要保证a1a1
即1.5
Sn + Sn-1 = 3n^2
Sn+1 + Sn = 3(n+1)^2
Sn+1 - Sn-1 = 3(n+1)^2 - 3n^2 = 6n+3
即 an+1 + an = 6n+3
an+2 + an+1 = 6(n+1)+3=6n+9
an+2 - an = 6
所以,任何情况下an < an+2,且 an+2 - an = 6
所以只要保证a1a1
即1.5