an=Sn-Sn-1=2an-2an-1-2^(n-1) (n>2)
an=2an-1+2^(n-1)
an/2^n=an-1/2^(n-1)+1/2
即bn=bn-1+1/2
bn为等差数列
s1=a1=2a1-2 a1=2
b1=a1/2=1
那么bn=b1+(n-1)2=(n+1)/2
an=bn*2^n=(n+1)*2^(n-1)
an=Sn-Sn-1=2an-2an-1-2^(n-1) (n>2)
an=2an-1+2^(n-1)
an/2^n=an-1/2^(n-1)+1/2
即bn=bn-1+1/2
bn为等差数列
s1=a1=2a1-2 a1=2
b1=a1/2=1
那么bn=b1+(n-1)2=(n+1)/2
an=bn*2^n=(n+1)*2^(n-1)