Question

已知a1=1,an+1=an+2^n，求an.

Answer 1

a1=1=2^0
a2=a1+2^1=2^0+2^1
a3=a2+2^2=2^0+2^1+2^2
a4=a3+2^3=2^0+2^1+2^2+2^3
∴an=2^0+2^1+2^2+2^3+……+2^(n-1)，为一等比数列，根据等比数列公式可知
an=1*(1-2^n)/(1-2)=2^n-1

Answer 2

已知 a_(n+1) = a_n + 2^n
得 a_(n+1) - a_n = 2^n
a_(n) - a_(n -1) = 2^(n-1)
a_(n-1) - a_(n-2) = 2^(n-2)
......
......
a_3 - a_2 = 2^2
a_2 - a_1 = 2^1
a_1 = 1
上式连加，得：
a_(n+1) = 1 + 2^1 + 2^2 +......+ 2^(n-2)+ 2^(n-1)+ 2^n
根据等比数列求和公式，得：
a_(n+1) = 2^(n+1)-1
∴ a_n = 2^n-1 (n>1)
a_1 = 1 满足公式
∴ a_n = 2^n-1