Question

Proszę o rozwiazanie zadania 9

Answer 1

a1 + a4 = 6 1/36 => a1 + a1 * q^3 = 6 1/36 => a1(1 + q^3) = 6 1/36 =>

a1 + a4 = 6 1/36 => a1 + a1 * q^3 = 6 1/36 => a1(1 + q^3) = 6 1/36 => a1 = (6 1/36)/(1+q^3)

a2 + a3 = 7/6 => a1 * q + a1 * q^2 = 7/6 => a1 * q(1 + q) = 7/6 =>

a2 + a3 = 7/6 => a1 * q + a1 * q^2 = 7/6 => a1 * q(1 + q) = 7/6 => a1 = (7/6)/(q(1+q))

a2 + a3 = 7/6 => a1 * q + a1 * q^2 = 7/6 => a1 * q(1 + q) = 7/6 => a1 = (7/6)/(q(1+q))(6 1/36)/(1+q^3) = (7/6)/(q(1+q)) // *(1+q^3) // *(q(1+q))

a2 + a3 = 7/6 => a1 * q + a1 * q^2 = 7/6 => a1 * q(1 + q) = 7/6 => a1 = (7/6)/(q(1+q))(6 1/36)/(1+q^3) = (7/6)/(q(1+q)) // *(1+q^3) // *(q(1+q))(6 1/36)(q(1+q)) = (7/6)(1+q^3) // *36

217q + 217q^2 = 42 + 42q^3

-42q^3 + 217q^2 + 217q - 42 = 0

tabela Hornera dla -1:

___-42__217__217__-42

___XX__42___-259_42

-1__-42__259_-42___0 - reszty

czyli:

-42q^3 + 217q^2 + 217q - 42 =

(q + 1)(-42q^2 + 259q - 42)

(-42q^2 + 259q - 42) - równanie kwadratowe, dla którego pierwiastki znajdziemy za pomocą delty: Δ = b^2 - 4ac = 259^2 - 4 * -42 * -42 =

67 081 - 7 056 = 60 025

√(60 025) = 245

q1 i q2 z wzoru (-b +- √Δ)/2a, to:

q1 = (-259 + 245)/-84 = -14/-84 = 1/6,

q2 = (-259 - 245)/-84 = -504/-84 = 6, więc:

(q + 1)(-42q^2 + 259q - 42) =

(q + 1)(q - 1/6)(q - 6)

=>   q = -1, q = 1/6, *q = 6*

tylko q = 6 da ciąg geometryczny rosnący (jak w poleceniu),

mnożenie przez q = -1 będzie powodować tylko zmianę znaku, a przez q = 1/6 otrzymany ciąg będzie malejący.

a1 = (7/6)/(6*7) = (7/6)/42 = 7/252 = 1/36

a1 = (7/6)/(6*7) = (7/6)/42 = 7/252 = 1/36a2 = 1/36 * 6 = 1/6

a1 = (7/6)/(6*7) = (7/6)/42 = 7/252 = 1/36a2 = 1/36 * 6 = 1/6a3 = 1/6 * 6 = 1

a1 = (7/6)/(6*7) = (7/6)/42 = 7/252 = 1/36a2 = 1/36 * 6 = 1/6a3 = 1/6 * 6 = 1a4 = 1 * 6 = 6