Maths : Surd

2014-07-13 8:18 pm
Simplify : (a + sqrt b)(c + sqrt d)^n + (a - sqrt b)(c - sqrt d)^n.
Answer in terms of a,b,c,d and n. (Note: a,b,c,d and n are positive integers.)

回答 (4)

2014-07-15 7:35 am
✔ 最佳答案
如果是 follow-up question, 應該問:
Simplify (a + b√x)(c + d√x)^n + (a - b√x)(c - d√x)^n

2014-07-14 09:56:49 補充:
多謝你那麼看起我,但問者不修正題目,我沒有理由答非所問的,

而且 Simplify (a + √b)(c + √d)^n + (a - √b)(c - √d)^n

並不是沒有可能的,只不過答案還是有根號且複雜,不像簡化。

2014-07-14 11:57:24 補充:
因為真的很複雜,又有√b,√d,√(bd),且可能看漏,太易錯了,不打算做。

若是我所説的題目,那反而可做一做。

2014-07-14 14:12:52 補充:
Let (c + d√x)^n = P + Q√x, so (c - d√x)^n = P - Q√x, where,
P = c^n + C(n,2)*c^(n-2)*d^2*x + ... + C*(n,2r)c^(n-2r)*d^(2r)*x^r + ...
and
Q = (√x)*[C(n,1)*c^(n-1)*d + C(n,3)*c^(n-3)*d^3*x + ... + C*(n,2r+1)c^(n-2r-1)*d^(2r+1)*x^r + ... ]

2014-07-14 14:13:30 補充:
So,
(a + b√x)(c + d√x)^n + (a - b√x)(c - d√x)^n
= (a + b√x)(P + Q√x) + (a - b√x)(P - Q√x)
= 2aP + 2bQ√x

2014-07-14 14:14:31 補充:
If a = -21, b = 14, c = 1/2, d = 1/2, x = 5, then,
when n = 1, the value is :
2*(-21)*(1/2) + 2*(14)*(√5)(1/2)*(√5)
= -21 + 70
= 49
When n = 2, the value is :
2*(-21)*[(1/2)^2 + 5(1/2)^2] + 2*(14)*(√5)[2(1/2)^2]*(√5)
= -63 + 70
= 7

2014-07-14 14:20:09 補充:
When n = 3, the value is :
2*(-21)*[(1/2)^3 + 3(5)(1/2)^3] + 2*(14)*(√5)[3(1/2)^3 + 5(1/2)^3]*(√5)
= -84 + 140
= 56
When n = 4, the value is :
2*(-21)*[(1/2)^4 + 6(5)(1/2)^4 + 25(1/2)^4] + 2*(14)*(√5)[4(1/2)^4 + 4(5)(1/2)^4]*(√5)
= -147 + 210
= 63

2014-07-14 23:35:37 補充:
感謝 知識長 的鼓勵,原來又不是想像中的複雜。

If T(n) = (a + √b)(c + √d)^n + (a - √b)(c - √d)^n
Simplify T(n)

Let (c + √d)^n = P + Q√d, so (c - √d)^n = P - Q√d, where,
P = c^n + C(n,2)*c^(n-2)*d + ... + C(n,2r)*c^(n-2r)*d^r + ...
and
Q = C(n,1)*c^(n-1) + C(n,3)*c^(n-3)*d + ... + C*(n,2r+1)c^(n-2r-1)*d^r + ...
So,
(a + √b)(c + √d)^n + (a - √b)(c - √d)^n
= (a + √b)(P + Q√d) + (a - √b)(P - Q√d)
= 2aP + 2Q√(bd)

As T(n) = 7(-3 + 2√5)[1/2 + (√5)/2]^n + 7(-3 - 2√5)[1/2 - (√5)/2]^n
Let a = -3, b = 20, c = 1/2, d = 5/4
When n = 1,
T(1)
= 7(2)(-3)(1/2) + 7(2)(1)√(20*5/4)
= -21 + 70
= 49
When n = 2,
T(2)
= 7(2)(-3)[(1/2)^2 + 5/4] + 7(2)[2(1/2)]√(20*5/4)
= -63 + 70
= 7
When n = 3,
T(3)
= 7(2)(-3)[(1/2)^3 + 3(1/2)(5/4)] + 7(2)[3(1/2)^2 + 5/4]√(20*5/4)
= -84 + 140
= 56
When n = 4,
T(4)
= 7(2)(-3)[(1/2)^4 + 6(1/2)^2(5/4) + (5/4)^2] + 7(2)[4(1/2)^3 + 4(1/2)(5/4)]√(20*5/4)
= -147 + 210
= 63
2014-07-20 9:30 pm
原式當然是最簡的,但由於程度問題,問者以為可以再簡單些。
2aP + 2Q√(bd)
其中
P = c^n + C(n,2)*c^(n-2)*d + ... + C(n,2r)*c^(n-2r)*d^r + ...
Q = C(n,1)*c^(n-1) + C(n,3)*c^(n-3)*d + ... + C*(n,2r+1)c^(n-2r-1)*d^r + ...
只是換了一個答法,當然沒有簡化,可能只是簡單表達了。
2014-07-16 5:58 am
2aP + 2Q√(bd)中的P及Q未能滿足以a,b,c,d,n表示答案的要求。
個人不相信這題是有解的。原式應該是最簡的了。
2014-07-14 2:56 am
Hahaha, this is clearly a follow-up question for me.

Thanks!

https://hk.knowledge. yahoo.com/question/question?qid=7014071300032

2014-07-13 19:35:21 補充:
http://www.wolframalpha.com/input/?i=49%2C7%2C56%2C63%2C119%2C182%2C301

T(n) = 7 ( 10 F(n) - 3 L(n) ) ?

2014-07-14 09:37:36 補充:
有道理~

其實我正想 email 你請你作答~

因為以我觀察,少年時 老師 一直以來都是 根號的專家。

2014-07-14 10:45:22 補充:
若你有空兩題並答,那則不是答非所問,而是提供更多的有用參考資料。

2014-07-17 16:06:21 補充:
謝謝 自由自在 ( 知識長 ),但發問者似乎也不太滿意我在上題的原式答案。

不打緊吧,盡了力就好啦~

如果大家將來見到可以化簡可以告訴我~

(。◕‿◕。)


收錄日期: 2021-04-23 23:25:46
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20140713000051KK00045

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