(1)当n=1时,有 a 1 = S 1 =
当n=2时,有 a 1 + a 2 =
当n≥2时,有 a n = S n - S n-1 =
∴(a n +a n-1 )(a n -a n-1 -2)=0又∵a n >0,∴a n -a n-1 =2, ∴数列{a n }是以1为首项,以2为公差的等差数列, ∴a n =1+(n-1)×2=2n-1 (2)由于b n =20-a n =21-2n,则b 1 =19,b n -b n-1 =-2<0. ∴{b n }是递减数列, 令
∴n=10,即数列{b n }的前10项和最大. |