Question

Oblicz sume ciagu geometrycznego:
a) -3+1-1/3+...-1/27
b) 8+4+2+...+1/16
c) 27+18+12+...+256/243

Answer 1

Potrzebne informacje:
an=a1*q^(n-1)
Sn=a1*(1-q^n)/(1-q) ^---potegowanie

a) -3+1-1/3+...-1/27
a1=-3
q=a2/a1=-1/3
ze wzoru na
an=-1/27 musze obliczyc n

-1/3³=-3*(-1/3)^(n-1)
-1/3³=-(1/3)^-1*(-1/3)^(n-1)
-1/3³=(-1/3)^n--------->n=3
Sn=-3(1+1/27)/(1+1/3)=-3(27+1)/(27+9)=-3*28/36=-28/12=-7/3

b) 8+4+2+...+1/16
a1=8
q=a2/a1=1/2
szukam n
1/16=8*(1/2)^(n-1)
(1/2)^4=(1/2)^-3 * (1/2)^(n-1)
(1/2)^4=(1/2)^(n-4)
4=n-4-------->n=8

Sn=8[1-(1/2)^8]/[1-1/2]
S8=8[1-1/16*1/16][1/2]
S8=8*255/256*2=255/16

c) 27+18+12+...+256/243
a1=27
q=18/27=2/3

szukam n

256/243=27 * (2/3)^(n-1)
2^8/3^5=3^3 * (2/3)^(n-1) dziele prze 3^3
(2/3)^8= (2/3)^(n-1)
8=n-1------>n=9

S9=27[1-(2/3]^9/(1/3)

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