(1)
It is given that N is a positive integer. Prove that
3[4^(N+1)] + 2[6^(N+1)] is a multiple of 12.
(2)
It is given that a-(1/a) = 2√6. Find the value of each of the following.
(a) a + (1/a)
(b) [1/(a^2)] - a^2
F.4 maths---number systems
2009-09-16 2:01 am
回答 (1)
2009-09-16 2:43 am
✔ 最佳答案
1) 3[4^(N+1)] + 2[6^(N+1)] = 3*4*(4^n) + 2*6*(6^n)
= 12(4^n) + 12(6^n)
= 12(4^n + 6^n) is a multiple of 12.
2)
(a) a + (1/a)
From a - 1/a = 2√6
a^2 + 1/a^2 - 2a(1/a) = 24
a^2 + 1/a^2 = 26
a^2 + 1/a^2 + 2a(1/a) = 26 + 2a(1/a)
(a + 1/a)^2 = 28
a + 1/a = 2√7 or - 2√7
(b) [1/(a^2)] - a^2
= (1/a - a)(1/a + a)
= (- 2√6)(2√7) or (- 2√6)(- 2√7)
= - 4√42 or 4√42
收錄日期: 2021-04-13 16:51:17
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