(1)an = -n^2 + 13n - 12 = -(n-1)(n-12)
所以当n < 12时,an >0,n = 12时,an = 0,n > 12时,an < 0.于是n = 12和11时,an的前n项和最大
(2)bn = Sn - S(n-1) = [√Sn + √S(n-1) ] * [√Sn - √S(n-1) ] = √Sn + √S(n-1),所以
√Sn - √S(n-1) = 1,而S1 = b1 = 1,所以Sn = n^2 ,bn = Sn - S(n-1) = n^2 - (n-1)^2 = 2n - 1
(3)最后一问cn的定义看不清楚.不过我没猜错的话,类似cn = 2^n * (2n-1),即通项是一个等比数列和等差数列的乘积.这应该是很常见的数列求和吧