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
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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
对一切n属于N*都成立证明结论
![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](/uploads/image/z/17538690-66-0.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E4%B9%8B%E5%92%8C%E4%B8%BASn%2C%E6%98%AF%E5%90%A6%E5%AD%98%E5%9C%A8%E5%B8%B8%E6%95%B0a%2Cb%2Cc%2C%E4%BD%BF%E5%BE%97an%3Dan%5E2%2Bbn%2Bc%2C%E6%BB%A1%E8%B6%B3a1%3D1%2C3Sn%3D%28n%2B2)
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!
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!
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