Question

已知数列{an}的前n项之和为Sn,是否存在常数a,b,c,使得an=an^2+bn+c,满足a1=1,3Sn=(n+2)an,

对一切n属于N*都成立证明结论

Answer 1

总体思路：
1）根据 a1 = 1 An = an^+ bn + c Sn = (n+2)An / 3

可列出 三个有关 a,b,c的方程，
S1 = (1+2)A1/3 = A1 得： a + b + c = 1
S2 = (2+2)A2/3 = A1 + A2 得： 4a + 2b + c = 3
S3 = (3+2)A3/3 = A1 + A2 + A3 得： 6a - c = 3

从而求得 a = 1/2 , b = 1/2 , c = 0

2) 这样就可得到 An = (n^+ n)/2 = n(n+1)/2
Sn = n(n+1)(n+2)/6

3) 当 n = 1, n =2, n=3 时 Sn = n(n+1)(n+2)/6 成立, 也即
3Sn=(n+2)An 成立！

现在只要当 n = k 时成立，n = k+1 时 Sn = n(n+1)(n+2)/6
也成立即可

假设 n = k 时 成立，则有 Sk = k(k+1)(k+2)/6

现在只要证 当n = k+1 时 S(k+1) = Sk + A(k+1) 则可

S(k+1) = (k+3)(k+1)(k+2)/6
Sk + A(k+1)= k(k+1)(k+2)/6 + (k+1)(k+2)/2

证明两者相等不难，且无论k取什么N*都可以相等
也即命题成立！