Răspuns:
1
Explicație pas cu pas:
a = (2/3)ⁿ : [(2ⁿ⁺¹+6ⁿ⁺¹)/(3ⁿ⁺¹+9ⁿ⁺¹)] ; n ∈ N
a = (2/3)ⁿ : [ (2ⁿ⁺¹ + 2ⁿ⁺¹·3ⁿ⁺¹)/(3ⁿ⁺¹+3ⁿ⁺¹·3ⁿ⁺¹)]
a = (2/3)ⁿ : {[2ⁿ⁺¹·(1+3ⁿ⁺¹)]/[3ⁿ⁺¹·(1+3ⁿ⁺¹)]}
a = (2/3)ⁿ : (2ⁿ⁺¹/3ⁿ⁺¹)
a = (2/3)ⁿ ·(3ⁿ⁺¹/2ⁿ⁺¹)
a = (2ⁿ·3·3ⁿ)/(3ⁿ·2·2ⁿ)
a = 3/2
(2a/3)¹⁰⁰ = [(2·3/2)/3]¹⁰⁰ = (3/3)¹⁰⁰ = 1¹⁰⁰ = 1
