Răspuns:
Folosim relatia: 1+a+a²+..+aⁿ = (1-aⁿ⁺¹)/(1-a)
a = -1/3
n = 2011
⇒ 1+(−1/3)+(-1/3)²+(-1/3)³+....+(−1/3)²⁰¹¹ = [1-(-1/3)²⁰¹²]/[1+1/3)
1+(−1/3)+(-1/3)²+(-1/3)³+....+(−1/3)²⁰¹¹ = (3/4)·[1-(1/3)²⁰¹²]
⇒ (1−1/3+1/3^2−1/3^3+....−1/3^2011):(1−1/3^2012) = (3/4)·[1-(1/3)²⁰¹²]/[1-(1/3)²⁰¹²] = 3/4
