10月15日尖子生练案50页:数列{an}满足[a(n+1)+an-1]/[a(n+1)-an+1]=n(n为正整数),
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10月15日尖子生练案50页:数列{an}满足[a(n+1)+an-1]/[a(n+1)-an+1]=n(n为正整数),且a2=6,则数列an的通
![](http://img.wesiedu.com/upload/d/26/d2667a53e3af8bd1926b7b722170eca1.jpg)
我是这么做的,看分子分母项,我试着弄了个负号,就是-n,那-n+n刚好等于0,分母肯定不为0,分子是2(an-2),这么算出来岂不是an=2哟,怎么回事?
![](http://img.wesiedu.com/upload/d/26/d2667a53e3af8bd1926b7b722170eca1.jpg)
我是这么做的,看分子分母项,我试着弄了个负号,就是-n,那-n+n刚好等于0,分母肯定不为0,分子是2(an-2),这么算出来岂不是an=2哟,怎么回事?
楼主英明,将a(n)=2带入已知条件,矛盾.因此,楼主的怀疑的对滴 .a(n)=2肯定有问题.
俺的解法如下:
a(n+1) + a(n) - 1 = na(n+1) - na(n) + n,
(n-1)a(n+1) = (n+1)a(n) - (n+1),
[先算出a(1)的值.]
(2-1)a(2) = 2a(1) - 2 = a(2) = 6, a(1)=4.
(n-1)a(n+1) = (n+1)a(n) - (n+1),[方程两边 准备同除(n-1)n(n+1), 但n=1时,(n-1)n(n+1)=0.所以,同除时,要保证n>=2.]
na(n+2) = (n+2)a(n+1) - (n+2), [这时,方程两边可以同除n(n+1)(n+2)了.]
a(n+2)/[(n+1)(n+2)] = a(n+1)/[n(n+1)] - 1/[n(n+1)] = a(n+1)/[n(n+1)] - 1/n + 1/(n+1),
a(n+2)/[(n+1)(n+2)] - 1/(n+1) = a(n+1)/[n(n+1)] - 1/n,
{a(n+1)/[n(n+1)] - 1/n}是首项 为a(2)/[1*2] - 1 = 2,的常数数列 .
a(n+1)/[n(n+1)] - 1/n = 2,
a(n+1)/[n(n+1)] = 2 + 1/n = (n+2)/n,
a(n+1) = n(n+1)(n+2)/n = (n+1)(n+2).
{a(n)}的通项公式为:
a(1)=4.
n>=2时,a(n) = n(n+1).
再问: 你好!问题是我前面几步哪里出现了问题啊?
再答: 楼主好。。您说“我试着弄了个负号,就是-n”
这是啥意思?
俺的解法如下:
a(n+1) + a(n) - 1 = na(n+1) - na(n) + n,
(n-1)a(n+1) = (n+1)a(n) - (n+1),
[先算出a(1)的值.]
(2-1)a(2) = 2a(1) - 2 = a(2) = 6, a(1)=4.
(n-1)a(n+1) = (n+1)a(n) - (n+1),[方程两边 准备同除(n-1)n(n+1), 但n=1时,(n-1)n(n+1)=0.所以,同除时,要保证n>=2.]
na(n+2) = (n+2)a(n+1) - (n+2), [这时,方程两边可以同除n(n+1)(n+2)了.]
a(n+2)/[(n+1)(n+2)] = a(n+1)/[n(n+1)] - 1/[n(n+1)] = a(n+1)/[n(n+1)] - 1/n + 1/(n+1),
a(n+2)/[(n+1)(n+2)] - 1/(n+1) = a(n+1)/[n(n+1)] - 1/n,
{a(n+1)/[n(n+1)] - 1/n}是首项 为a(2)/[1*2] - 1 = 2,的常数数列 .
a(n+1)/[n(n+1)] - 1/n = 2,
a(n+1)/[n(n+1)] = 2 + 1/n = (n+2)/n,
a(n+1) = n(n+1)(n+2)/n = (n+1)(n+2).
{a(n)}的通项公式为:
a(1)=4.
n>=2时,a(n) = n(n+1).
再问: 你好!问题是我前面几步哪里出现了问题啊?
再答: 楼主好。。您说“我试着弄了个负号,就是-n”
这是啥意思?
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