Question

Determine a soma dos termos da P.G. (1,3,9 ...,729 )

Answer 1

3^0 + 3^1+.....+3^6 = (3^7 -1)/2 = 2186/2 = 1093

Answer 2

An = 729

a₁ = 1

q = a₂ / a₁

q = 3/1

q = 3

n = ?

An = a₁ * qⁿ⁻¹

729 = 1 * 3ⁿ⁻¹

3ⁿ⁻¹ = 729 / 1

3ⁿ⁻¹ = 729

3ⁿ⁻¹ = 3⁶

n - 1 = 6

n = 1 + 6  

n = 7

Sn = a₁ * (qⁿ - 1) / q - 1

S₇ = 1 * (3⁷ - 1) / 3 - 1

S₇ = 2187 - 1 / 2

S₇ = 2186 / 2

S₇ = 1 093