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2010-02-10T19:08:56+01:00
Lewa strona przedstawia sume wyrazow ciagu geometrycznego

a₁=1/3
q=1/3
S(n)=1/3 * (1-(1/3)^n)/(1-1/3) = 364/729
(1-(1/3)^n)/(2/3) = 364/243
1-(1/3)^n = 364/243 * 2/3
(1/3)^n = 1 - 728/729
(1/3)^n = 1/729
(1/3)^n = (1/3)^6
n = 6
x - 6 wyraz tego ciagu
x = a₁*q⁵ = 1/3 * (1/3)^5 = (1/3)^6 = 1/729
2 5 2
2010-02-10T20:00:46+01:00
X=364/729-1/3 - 1/9 -1/27-... <1/3>
1/3 * (1-(1/3)^n)/(1-1/3) = 364/729
(1/3)^n = 1/729
1/3 * (1/3)^5 = (1/3)^6 = 1/729
1 1 1
2010-02-10T23:03:18+01:00
1/3 + 1/9 +1/27+...+x=364/729
a₁=1/3
q= 1/9*3/1= 1/3
ciąg geometryczny

1/3* ((1-(1/3)^n/ (1-1/3) )= 367/729
1/3*(1-(1/3^n)*(3/2) =364/729
1/2*(1-1/3^n) =364/729 *2
1-(1/3)^n = 728/729
-(1/3)^n = 729/729 -1
1/3^n= 1/729
1/729 = (1/3)⁶
n=6

ciąg ten ma 6 wyrazów ostatni to x

x₆ = (1/3)⁶ = 1/729

1 1 1