calcule a soma dos 12 primeiros termos da PG dada abaixo para cada um dos valores de m indicados. (3m,3m²,3m³..)
a)m=1 b) m= -2 c) m=3 d) m= 1/2
Respostas
a) m=1
PG(3,3,3,3,3,3,3,3,3,3,3,3)
Soma=12x3=36
b) m=-2
PG(-6, 12, -24...)
calculando a12
:
Agora calculando a soma:
c)
PG(9,27, 81...)
calculando a12:
Agora calculando a soma:
d)
PG(3/2, 3/4, 3/8...)
q=1/2
calculando a12:
Agora calculando a soma:
a)
m=1 => PG(3,3,3,3,3,3,3,3,3,3,3,3)
S12=12x3=36
b) m=-2 => (-6, 12, -24...)
a12= (-6).(-2)^(12-1)
a12= (-6).(-2)^11 => -6(-2048) => 12288
S12 = (-6)(-(-2)^12
1-(-2)
S12=
c)(9,27, 81...)
a12=9.(3)^(12-1)
a12=9.3^11
a12= 9. 177147 => a12 = 1594323
S12= 9.(1-(3)^12)
1- 3
S12= 9.(1 -3^12)
- 2
S12= 2391480
d) (3/2, 3/4, 3/8...)
a12=3/2.(1/2)^(12-1)
a12=3/2(1/2)^11
a12=3/2. 1/2048
a12= 3/4096
S12 = (3/2)(1-(1/2)^12)
1-(1/2)
s12= 3/2(1- 1/2048) => 3/2.( 2048-1)/2048 => 3 . 2047 . 2 => 6141
( 2-1)/2 1/2 2. 2048 1 2048