对于数列{an},定义数列{a(n+1)-an}为数列{an}的差数列,若a1=2

{an}的差数列的通项为2^n所以a(n+1)-an=2^nan-a(n-1)=2^(n-1).a3-a2=2^2a2-a1=2^1将上述式子相加a(n+1)-a1=2^n+2^(n-1)+...+2^2+2^1=2^(n+1)-2a1=2an=2^nSn=a1(1-q^n)/(1-q)=2(1-2^n)/(1-2)=2^(n+1)-2...