对于无穷数列Sn和函数fx若Xn+1=fxn则称fx为数列Xn的母函数.求证fx=x+2是数列2的n次方乘an的母函数.求an的前n项和

Question

对于无穷数列Sn和函数fx若Xn+1=fxn则称fx为数列Xn的母函数.求证fx=x+2是数列2的n次方乘an的母函数.求an
的前n项和

Answer 1

f(x)=x+2,
b(n)=2^na(n),
b(n+1) = f[b(n)] = b(n) + 2,
{b(n)=2^na(n)} 是首项为b(1)=2a(1),公差为2的等差数列.
2^na(n) = 2a(1)+2(n-1),
a(n) = [a(1)+(n-1)]/2^(n-1) ,
s(n) = a(1)/1 + [a(1)+1]/2 + [a(1)+2]/2^2 + ...+ [a(1)+n-2]/2^(n-2) + [a(1)+n-1]/2^(n-1),
2s(n) = 2a(1)+[a(1)+1]/1 + [a(1)+2]/2+[a(1)+3]/2^2+...+[a(1)+n-2]/2^(n-3)+ [a(1)+n-1]/2^(n-2),
s(n) = 2s(n)-s(n)
=2a(1)+1 + 1/2 + 1/2^2 + ...+ 1/2^(n-2) - [a(1)+n-1]/2^(n-1)
=2a(1) + [1-1/2^(n-1)]/[1-1/2] - [a(1)+n-1]/2^(n-1)
=2a(1) + 2-2/2^(n-1) - [a(1)+n-1]/2^(n-1)
=2a(1) + 2 - [a(1)+n+1]/2^(n-1)