Question

数列{an}中,a1=1,an+1=an+2^n,则通项公式为an=

Answer 1

∵an+1=an+2^n
∴an=a(n-1)+2^(n-1)
a(n-1)=a(n-2)+2^(n-2)
a(n-2)=a(n-3)+2^(n-2)
.........
a2=a1+2
将以上式子加起来得an+a(n-1)+....a2=a(n-1)+a(n-2)+....a1+2^(n-1)+....+2¹
an=a1+2¹+2²+.....2^(n-1)
an=2º+2¹+2²+.....2^(n-1)=2^n-1

Answer 2

an-an-1=2^n-1;
an-1-an-2=2^n-2;
......
......
a3-a2=2^2;
a2-a1=2^1;
以上各式相加左边只剩下an-a1，得：
an-a1=2^1+2^1+.....+2^n-1
=2*(2^n-1-1)/(2-1)=2^n-2
又因为a1=3，所以an=2^n-2+3=2^n+1
检验a1也符合该式，所以an=2^n+1