polecenie: rozwiaz równanie: 1/3 + 1/9 +1/27+...+x=364/729 P.S. radze nie spamowac przy tym zadaniu!!

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

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

1/3 + 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