Question

在数列an中,a1=1,an+1 2an+2的n次方

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