The document proves that for any natural number n ≥ 2, the sum of the reciprocals of the square roots of the first n natural numbers is greater than the square root of n. It shows that if this property holds for some number k, it also holds for k+1. Therefore, by mathematical induction, the property is true for any natural number n ≥ 2. The document also proves that the sum of the first n terms of the geometric progression 3, 9, 27, ... is equal to (3/2)(3n - 1) by using a similar induction argument.

1. a) Demuestra que para todo número natural n ≥ 2
𝟏
√𝟏
+
𝟏
√𝟐
+
𝟏
√𝟑
+ . .. +
𝟏
√𝒏
> √𝒏
Sea 𝑃(𝑛): √𝑛 <
1
√1
+
1
√2
+
1
√3
+ ⋯
1
√𝑛
, ∀ ℕ 𝑛 ≥ 2
Para 𝑛 = 2
𝑃(2):√𝑛 <
1
√1
+
1
√2
𝑃(𝑛) es cierta para 𝑛 = 𝑘
𝑃(𝑘): √𝑘 <
1
√1
+
1
√2
+ ⋯+
1
√𝑘
Para probar que 𝑃(𝑘 + 1) es cierta, demostramos que:
𝑃(𝑘 + 1):√𝐾 + 1 <
1
√1
+
1
√2
+ ⋯+
1
√𝑘
+
1
√𝑘 + 1
1
√1
+
1
√2
+ ⋯ +
1
√𝑘
+
1
√𝑘 + 1
> √𝑘 +
1
√𝑘 + 1
> √𝑘 + 1 ∴
1
√𝑘+1
> 0
De aquí, si la fórmula es cierta para 𝑃(𝑘) se ha demostrado que también se verifica para el siguiente,
𝑃(𝐾 + 1)
Por lo tanto, 𝑃(𝑛) es válido para cualquier número natural n, 𝑛 ≥ 2
b) 𝟑 + 𝟗 + 𝟐𝟕 + ⋯ + 𝟑𝒏
= (
𝟑
𝟐
)( 𝟑𝒏
− 𝟏 )
1. 𝑛 = 1
3 =
3
2
(31 − 1)
3 =
3
2
(2) =
6
2
= 3
2. 𝑛 = 𝑘
3 + 9 + 27 + ⋯+ 3𝑘−1
= (
3
2
)( 3𝑘
− 1 )
3. 𝑛 = 𝑘 + 1
3 + 9 + 27 + ⋯+ 3𝑘−1
+ 3𝑘
= (
3
2
)( 3𝑘+1
− 1)
3
2
(3𝑘
− 1) + 3𝑘
=
3
2
(3𝑘+1
− 1)

2. 3 ∙ 3𝑘+1
− 1
2
+
3𝑘
2
= (
3
2
)( 3𝑘
− 1 )