Question

数学归纳法求证1+r+r^2+...+r^n=[(r^n+1)-1]/r-1

Answer 1

证：

n = 0

{[(r^(0+1)]-1}/(r-1) = 1

若对n = a,a≥0有

1+r+r^2+...+r^a={[(r^(a+1)]-1}/(r-1)

n = a+1

1+r+r^2+...+r^a+r^(a+1)

= {[(r^(a+1)]-1}/(r-1) +r^(a+1)

= {[(r^(a+1)]-1 + (r-1) * r^(a+1)}/(r-1)

= {[(1+r-1) * (r^(a+1)]-1}/(r-1)

= {r^[(a+1)+1] - 1} / (r-1)

则对n = a+1,1+r+r^2+...+r^a+r^(a+1)= {r^[(a+1)+1] - 1} / (r-1)也成立

综上所述,由数学归纳法得知,对任意自然数n,1+r+r^2+...+r^n=[(r^n+1)-1]/r-1.