Question

在数列an中，a1=1，an+1 2an+2的n次方

1.设bn=an/2的n-1次方，证明：数列bn是等差数列
2求数列an的前n项和Sn

Answer 1

a(n+1)=2an+2^n,bn=an/2^(n-1),b(n+1)=a(n+1)/2^n,b1=a1/2^0=1
a(n+1)/2^n=an/2^(n-1)+1,b(n+1)=bn+1,bn为首项为1公差为1的等比数列，bn=n
an=n*2^(n-1),a1=1,s1=1
sn-s(n-1)=n*2^n
sn/2^n-s(n-1)/2^n=n
sn/2^n-1/2*s(n-1)/2^(n-1)=n
sn/2^n=1/2(s(n-1)/2^(n-1)))+n
sn/2^n-2n=1/2*((s(n-1)/2^(n-1)))-2n)
令sn/2^n-2n=cn,c1=1/2-2=-3/2,cn=1/2c(n-1),
cn=(1/2)^(n-1)c1=-3/2(1/2)^(n-1)=-3(1/2)^n=-3*n^(-n)
sn/2^n-2n=-3*2^(-n)
sn/2^n=-3*2^(-n)+2n
sn=-3+2n*2^n
sn=n*2^(n+1)-3