当n>1时
S(n) = a(n+1)
S(n-1) = a(n)
两式相减
S(n) - S(n-1) = a(n+1) - a(n)
左边显然 = a(n)
所以a(n) = a(n+1) - a(n)
a(n+1) = 2a(n)
S1 = a1 = 2,又S1 = a(1+1) = a(2) = 2
故a(n) = a(2)*2^(n-2) = 2^(n-1),n>=2
a(1) = 2
S(n) = a(n+1) = 2^n
这种类型的题(给出Sn和某项的关系),通常都是这样做的