2的N次方的求和怎么算?

Question

1+4+8+18+................2^n=?

Answer 1

设S=2^0+2^1+2^2+……+2^n则2S=2^1+2^2+……+2^(n+1)所以S=2S-S=(2^1+2^2+……+2^(n+1))-(2^0+2^1+2^2+……+2^n)=2^(n+1)-2^0=2^(n+1)-1

Answer 2

等比数列的求和，a1=1，q=2，有n+1项

Answer 3

Sn=a1(1-q^n)/(1-q);
a1:首项;
q:公比
n:项数