Question

Qual a soma dos termos da PG (3,12,...,12 288)

Answer 1

Q = a₂/a₁
Q = 12/3
Q = 4

$A_{n} = a_{1} . Q^{(n -1)}$
12288/3 = 4⁽ⁿ ⁻¹⁾
4096 = 4⁽ⁿ ⁻¹⁾
4⁶ = 4⁽ⁿ ⁻¹⁾

6 + 1 = n
7 = n

$S_{n} = \frac{a_{1} . (q^n -1) }{q -1}$
s7 = 3 . (4⁷ -1)/4 -1
s7 = 3 . (16384 -1)/3
s7 = 16383