10, x, 20 - początkowe wyrazy ciągu geometrycznego, zatem x /10 = 20/x x² = 10*20 = 200 x =√200 = √100*√2 = 10√2 Odp. x = 10√2 a1 = 10, a2 = 10√2, a3 = 20 q = a2 : a1 = a3 : a2 = 10√2 : 10 = √2 S16 = a1*[1 - q¹⁶]/[ 1 - q] = 10*[1 - (√2)¹⁶]/[1 - √2] = 10*[1 - 2⁸]/[1 - √2] = = 10*[1 - 256]/[1 -√2] = - 10*255/[1 -√2] = - 2550/[1 - √2] = = [ - 2 550 *(1 + √2)]/[1 -√2)(1 + √2)] = [ -2550(1 + √2)]/[1 -2] = = 2550 (1 + √2) ================================ (√2)¹⁶ = [(√2)²]⁸ = 2⁸ = 256
z ciągu geometrycznego: [latex]a_2^2 = (a_1q)^2 = a_1 * a_1q^2 = a_1*a_3\ x^2 = 10*20 = 200\ x = 10sqrt{2} vee x = - 10sqrt{2} [/latex] ale x > 0 iloraz: [latex]a_2 = a_1*q Rightarrow q = frac{a_2}{a_1}\ x = 10sqrt{2} Rightarrow q = frac{10sqrt{2}}{10} = sqrt{2}[/latex] ponieważ mamy x ≠ 1 to zachodzi: [latex]S_n = a_1frac{1 - q^n}{1 - q}\ q = sqrt{2} Rightarrow S_{16} = 10frac{1 - (sqrt{2})^{16}}{1 - sqrt{2}} = 10frac{2^8 - 1}{sqrt{2} - 1} = 10frac{(2^8 - 1)(sqrt{2} + 1)}{(sqrt{2} - 1)(sqrt{2} + 1)} = 10frac{(256 - 1)(sqrt{2} + 1)}{2 - 1} = 2550(sqrt{2} + 1) [/latex] jak masz pytania to pisz na pw
Iloraz ciągu geom.: q = a2/a1 = a3/a2 Czyli x/10 = 20/x x² = 200 x1 = 10 √2 (x2 = -10 √2 < 0 -- odrzucamy: wyrazy mają być dodatnie). q = x/a1 czyli: q = 10 √2 / 10 q = √2 Mamy więc ciąg: 10; 10√2; 20; 20√2; 40; . . . Suma: S = a1 * (q^n - 1) / (q - 1) S = 10 * ((√2)^16 - 1) / (√2 - 1) S = 10 * (2^8 - 1) / (√2 - 1) S = 10 * 255 /(√2 - 1) S = 2550 * (√2 + 1) / (2 - 1) S = 2550 * (√2 + 1)
