Prosiłbym do poniedziałku, naprawde jest mi to potrzebne, dzięki.... Question from @tomm

Janek191

z.1

a) an = (3n - 2) / 2^n

a1 = (3*1 - 2)/2^1 = 1/2

a2 =(3*2 - 2)/2^2 = 4/4 = 1

a3 = (3*3 - 2)/2^3 = 7/8

a4 = (3*4 - 2 )/2^4 = 10/16 = 5/8

a5 = (3*5 - 2)/2^5 = 13/32

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an = 4 - n^2

a1 = 4 - 1^2 = 4 - 1 = 3

a2 = 4 - 2^2 = 4 - 4 = 0

a3 = 4 - 3^2 = 4 - 9 = - 5

a4= 4 - 4^2 = 4 - 16 = - 12

a5 = 4 - 5^2 = 4 - 25 = - 21

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an = (-1)^n *(2n + 3)

a1 = (-1)^1 *(2*1+ 3) = - 5

a2 = (-1)^2 *(2*2 + 3) = 7

a3 = (-1)^3 *( 2*3 + 3) = - 9

a4 = (-1)^4 *(2*4 + 3) = 11

a5 = (-1)^5 *(2*5 + 3) = - 13

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z.11

Ciąg geometryczny

a) a3 = 6 oraz a4 = 18

Mamy

q = a4 : a3 = 18/6 = 3

a3 = a1*q^2 ---> a1 = a3 : q^2 = 6 : 3^2 = 6 : 9 = 2/3

Odp. a1 = 2/3 oraz q = 3

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a2 = 8 oraz a4 = 32

Mamy

a2 = a1*q oraz a4 = a1*q^3

zatem

a4 : a2 = a1*q^3 : a1*q = q^2

ale a4 : a2 = 32 : 8 = 4

zatem

q^2 = 4

czyli q = - 2 lub q = 2

a1 = a2 : q = 8 : (-2) = - 4

lub

a1 = a2: q = 8 : 2 = 4

Odp.

a1 = - 4 oraz q = -2

lub

a1 = 4 oraz q = 2

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z.12

q = - 2/3

a6 = 32/27

Mamy

a6 = a1*q^5 --> a1 = a6 : q^5

a1 = (32/27) : ( -2/3)^5 = (32/27) : ( - 32/243) = (32/27)*( - 243/32) =

= - 243/27 = - 9

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z.13

a1 = 3 oraz q = - 2

Oblicz

a) S6 = a1* [1 - q^6]/[1 - q]

S6 = 3*[ 1 - (-2)^6]/[1 - (-2)] = 3*[ 1 - 64]/3 = - 63

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S7 = a1*[1 - q^7]/[ 1 - q] = 3*[ 1 - (-2)^7]/[ 1 - (-2)] =

= 3*[ 1 - 128]/3 = 1 - 128 = - 127

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Potęgi i logarytmy

z.1

( 2 1/3)^(-3) * ( 6/7)^(-3) = ( 7/3 *6/7)^(-3) = 2^(-3) = 1/2^3 = 1/8

(1/9)^7 * 3 ^15 = (1/9)^7 * 3^7 * 3^8 = (1/9 * 3)^7 * 3^8 =

= (1/3)^7 *3^7 * 3 = 1*3 = 3

lub

(1/9)^7 * 3 ^15 = 3^15 / (3^2)^7 = 3^15/ 3^14 = 3

( 1 2/3)^(-2/3) * ( 16 1/5) ^(-2/3) =[(1/27)^(1/3)] ^2 =

= (1/3)^2 = 1/9

25^7 / 125^5 = [( 5^2)^7]/ [ (5^3)^5] = 5^14 / 5^15 = 1/5

[ 4^(1/3) / 8 ]^(-6/7) = [ 2^(2/3)/ 2^3]^( -6/7) = [ 2 ^(-7/3)]^(-6/7) =

= 2^((-7/3)*(-6/7)) = 2^2 = 4

===========================

z.2

log 3 [ 1/27 ] = - 3

log 5 [ 5^(1/3)] = 1/3

log 10^30 = 30

8^log 2 [1/5] = [2^3]^log 2 [ 1/5] = 2^ 3 log 2[1/5] = 2^log 2 [ (1/5)^3] =

= 2^log 2 [ 1/125] = 1/125

log 5[ 8 1/3] + log 5 [ 15] = log 5 [(25/3 ) * 15] = log 5 [ 125] = 3

log 0,004 - log 0,04 = log [ 0,004/ 0,04] = log [ 1/10] = - 1

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Roma

Zad.1

$a_n =\frac{3n - 2}{2^n}$

$a_1 =\frac{3\cdot 1 - 2}{2^1} = \frac{3 - 2}{2} = \frac{1}{2}$

$a_2 =\frac{3\cdot 2 - 2}{2^2} = \frac{6 - 2}{4} = \frac{4}{4} = 1$

$a_3 =\frac{3\cdot 3 - 2}{2^3} = \frac{9 - 2}{8} = \frac{7}{8}$

$a_4 =\frac{3\cdot 4 - 2}{2^4} = \frac{12 - 2}{16} = \frac{10}{16} = \frac{5}{8}$

$a_5 =\frac{3\cdot 5 - 2}{2^5} = \frac{15 - 2}{32} = \frac{13}{32}$

$a_n = 4 - n^2$

$a_1 = 4 - 1^2 = 4 - 1 = 3$

$a_2 = 4 - 2^2 = 4 - 4 = 0$

$a_3 = 4 - 3^2 = 4 - 9 = -5$

$a_4 = 4 - 4^2 = 4 - 16 = -12$

$a_5 = 4 - 5^2 = 4 - 25 = -21$

$a_n = (-1)^n \cdot (2n + 3)$

$a_1 = (-1)^1 \cdot (2 \cdot 1 + 3) = - 1 \cdot (2 + 3) = - 1 \cdot 5 = - 5$

$a_2 = (-1)^2 \cdot (2 \cdot 2 + 3) = 1 \cdot (4 + 3) = 1 \cdot 7 = 7$

$a_3 = (-1)^3 \cdot (2 \cdot 3 + 3) = - 1 \cdot (6 + 3) = - 1 \cdot 9 = - 9$

$a_4 = (-1)^4 \cdot (2 \cdot 4 + 3) = 1 \cdot (8 + 3) = 1 \cdot 11 = 11$

$a_5 = (-1)^5 \cdot (2 \cdot 5 + 3) = - 1 \cdot (10 + 3) = - 1 \cdot 13 = - 13$

Zad.11 Ciąg geometryczny

n-ty wyraz ciągu geometrycznego (an) o ilprazie q dla n ∈ N₊ \ {1} okreslony jest wzorem:

$a_n =a_1 \cdot q^{n-1}$

a) a₃ = 6 i a₄ = 18

a₄ = a₃ · q

q = a₄ : a₃ =

q = 18 : 6 = 3

a₃ = a₁ · q²

a₁ · 3² = 6

a₁ · 9 = 6 /:9

$a_1 =\frac{6}{9}$

$a_1 =\frac{2}{3}$

Odp. a₁ = ⅔ ; q = 3

a₂ = 8 i a₄ = 32

a₂ = a₁ · q

a₁ · q = 8

a₄ = a₁ · q³

a₁ · q³ = 32

$\left \{ {{a_1 \cdot q = 8 \ / : \ q} \atop {a_1 \cdot q^3 = 32}} \right$

$\left \{ {{a_1 = \frac{8}{q}} \atop {\frac{8}{q} \cdot q^3 = 32}} \right$

$\left \{ {{a_1 = \frac{8}{q}} \atop {8 \cdot q^2 = 32 \ / : \ 8}} \right$

$\left \{ {{a_1 = \frac{8}{q}} \atop {q^2 = 4}} \right$

$\left \{ {{a_1 = \frac{8}{q}} \atop {q = 2}} \right \ lub \ \left \{ {{a_1 = \frac{8}{q}} \atop {q = -2}} \right$

$\left \{ {{a_1 = \frac{8}{2}} \atop {q = 2}} \right \ lub \ \left \{ {{a_1 = \frac{8}{-2}} \atop {q = -2}} \right$

$\left \{ {{a_1 = 4} \atop {q = 2}} \right \ lub \ \left \{ {{a_1 = -4} \atop {q = -2}} \right$

Odp. a₁ = 4 i q = 2 lub a₁ = -4 i q = - 2

Zad. 12

n-ty wyraz ciągu geometrycznego (an) o ilprazie q dla n ∈ N₊ \ {1} okreslony jest wzorem:

$a_n =a_1 \cdot q^{n-1}$

q = - ⅔

a₆ = ³²/₂₇

a₆ = a₁ · q⁵

$\frac{32}{27} = a_1 \cdot (-\frac{2}{3})^5$

$\frac{32}{27} = a_1 \cdot (-\frac{32}{243}) \ / : \ (-\frac{32}{243})$

$a_1 = \frac{32}{27} \cdot (-\frac{243}{32})$

$a_1 = 9$

Zad. 13

Suma n początkowych wyrazów ciągu geoemtrycznego (an) wyraża się wzorem:

$S_n = \frac{a_1 \cdot (1 - q^n)}{1 - q}$

a₁ = 3 i q = - 2

$S_6 = \frac{3 \cdot [1 - (-2)^6]}{1 - (-2)} = \frac{3 \cdot (1 - 64)}{1+2} = \frac{3 \cdot (- 63)}{3} = -63$

$S_7 = \frac{3 \cdot [1 - (-2)^7]}{1 - (-2)} = \frac{3 \cdot [1 - (-128)]}{1+2} = \frac{3 \cdot (1+128)}{3} = 129$

Potęgi i logarytmy

Zad. 1

$( 2 \frac{1}{3})^{-3} \cdot ( \frac{6}{7})^{-3} = ( \frac{7}{3})^{-3} \cdot ( \frac{6}{7})^{-3} =( \frac{7}{3} \cdot \frac{6}{7})^{-3}= 2^{-3} = \frac{1}{2^3} = \frac{1}{8}$

$(\frac{1}{9})^7 \cdot 3^{15} = (\frac{1}{3^2})^7 \cdot 3^{15} = (3^{-2})^7 \cdot 3^{15} = 3^{-14} \cdot 3^{15} = 3^{-14+15} = 3^1 = 3$

$(1\frac{2}{3})^{-\frac{2}{3}} \cdot (16\frac{1}{5})^{-\frac{2}{3}} = (\frac{5}{3})^{-\frac{2}{3}} \cdot (\frac{81}{5})^{-\frac{2}{3}} =(\frac{5}{3} \cdot \frac{81}{5})^{-\frac{2}{3}} =$

$=27^{-\frac{2}{3}} = (3^3)^{-\frac{2}{3}} =3^{3 \cdot (-\frac{2}{3})} = 3^{-2} = \frac{1}{3^2} = \frac{1}{9}$

$\frac{25^7}{125^5} = \frac{(5^2)^7}{(5^3)^5} = \frac{5^{14}}{5^{15}} = 5^{14-15} = 5^{-1} = \frac{1}{5^1}= \frac{1}{5}$

$(\frac{\sqrt[3]{4}}{8})^{-\frac{6}{7}} = (\frac{4^{\frac{1}{3}}}{2^3})^{-\frac{6}{7}} = (\frac{(2^2)^{\frac{1}{3}}}{2^3})^{-\frac{6}{7}} = (\frac{2^{\frac{2}{3}}}{2^3})^{-\frac{6}{7}} =$

$=(2^{\frac{2}{3}-3})^{-\frac{6}{7}} = (2^{\frac{2}{3}-\frac{9}{3}})^{-\frac{6}{7}} = (2^{-\frac{7}{3}})^{-\frac{6}{7}} =2^{-\frac{7}{3}\cdot (-\frac{6}{7})} = 2^2 = 4$