Question

已知数列an满足a1=1,an+1=1/2an+1/2^n,求an的通项公式。

急急急~~~~~~高分求高手破解~~~~~~今晚10点以前~~~~~~~~

Answer 1

a[n+1]=a[n]/2+1/2^n
a[n]/2=a[n-1]/2^2+1/2^n
a[n-1]/2^2=a[n-2]/2^3+1/2^n
a[n-2]/2^3=a[n-3]/2^4+1/2^n
...
a[2]/2^(n-1)=a[1]/2^n+1/2^n

以上各式相加，两边相同的项消去可得：
a[n+1]=a[1]/2^n+n/2^n=(n+1)/2^n
于是a[n]通项公式为：a[n]=n/2^(n-1)，n是任意正整数.

Answer 2

递推式等号两边同时乘以2^(n+1)
得到
a(n+1)*2^(n+1)=a(n)*2^n+2
令
b(n)=a(n)*2^n
则数列b(n)是以b1=a1*2为首项,2为公差的等差数列