a+b=1,ab=-1,a7次方+b7次方
答案:2 悬赏:80
解决时间 2021-02-22 14:18
- 提问者网友：泪姬迷茫
- 2021-02-22 00:19
a+b=1,ab=-1,a7次方+b7次方
最佳答案
- 二级知识专家网友：开心就好
- 2021-02-22 00:32
a + b = 1，ab = - 1
S₁ = a + b = 1
S₂ = a² + b² = (a + b)² - 2ab = (1) - 2(- 1) = 3
S₃ = a³ + b³ = (a + b)(a² + b² - ab ) = (1)(3 - (- 1)) = 4
S₄ = a⁴ + b⁴ = (a²)² + (b²)² + 2a²b² - 2a²b² = (a² + b²)² - 2(ab)² = 3² - 2(- 1)² = 7
S(n) = aⁿ + bⁿ
S(n - 1) = aⁿ⁻¹ + bⁿ⁻¹
S(n - 2) = aⁿ⁻² + bⁿ⁻²
(a + b) · S(n - 1)
= (a + b) · (aⁿ⁻¹ + bⁿ⁻¹)
= aⁿ + bⁿ + ab(aⁿ⁻² + bⁿ⁻²)
= S(n) + ab · S(n - 2)
S(n - 1) = S(n) - S(n - 2) <==将a + b = 1，ab = - 1代入
S(n) = S(n - 1) + S(n - 2)
a⁷ + b⁷ = S(7)
= S(6) + S(5)
= [S(5) + S(4)] + [S(4) + S(3)]
= [S(4) + S(3) + S(4)] + [S(4) + S(3)]
= (7 + 4 + 7) + (7 + 4)
= 29
S₁ = a + b = 1
S₂ = a² + b² = (a + b)² - 2ab = (1) - 2(- 1) = 3
S₃ = a³ + b³ = (a + b)(a² + b² - ab ) = (1)(3 - (- 1)) = 4
S₄ = a⁴ + b⁴ = (a²)² + (b²)² + 2a²b² - 2a²b² = (a² + b²)² - 2(ab)² = 3² - 2(- 1)² = 7
S(n) = aⁿ + bⁿ
S(n - 1) = aⁿ⁻¹ + bⁿ⁻¹
S(n - 2) = aⁿ⁻² + bⁿ⁻²
(a + b) · S(n - 1)
= (a + b) · (aⁿ⁻¹ + bⁿ⁻¹)
= aⁿ + bⁿ + ab(aⁿ⁻² + bⁿ⁻²)
= S(n) + ab · S(n - 2)
S(n - 1) = S(n) - S(n - 2) <==将a + b = 1，ab = - 1代入
S(n) = S(n - 1) + S(n - 2)
a⁷ + b⁷ = S(7)
= S(6) + S(5)
= [S(5) + S(4)] + [S(4) + S(3)]
= [S(4) + S(3) + S(4)] + [S(4) + S(3)]
= (7 + 4 + 7) + (7 + 4)
= 29
全部回答
- 1楼网友：甜野猫
- 2021-02-22 01:03
(a+d)^2等a^2+2ab+b^2等1,所以ab等-1/2,(a+b)^3等a^3+3a^2b+3ab^2+b^3等1,所以a^3+b^3等5/2,依次算出a^4+b^4,a^5+b^5,a^6+b^6,a^7+b^7
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