Quod Erat Demonstrandum

2009/01/02

Something about F.6 Pure Math First Term Exam

Filed under: HKALE,Pure Mathematics — johnmayhk @ 4:24 下午
Tags: ,

Here is one of questions:

Given that

x_1 = 4, x_2 = 12
x_{n + 2} = 4(x_{n + 1} - x_{n}) (\forall n \in \mathbb{N})

Prove that

(a) x_n = 2(1 + \frac{1}{n})x_{n - 1}
(b) x_n = (n + 1)2^n

The question requires students to use mathematical induction to prove that.

I’d like to give another ways.

(method 1)

By looking at x_{n + 2} - 4x_{n + 1} + 4x_{n} = 0, we can solve x_n just like what we do in solving second order linear differential equation. What? Urm, the characteristic equation is

\lambda^2 - 4\lambda + 4 = 0

Yield, \lambda = 2 (repeated)

Hence,

x_n = (An + B)(2^n)

With initial conditions x_1 = 4 and x_2 = 12, we have

2(A + B) = 4 \Rightarrow A + B = 2
4(2A + B) = 12 \Rightarrow 2A + B = 3

On solving, A = B = 1, thus

x_n = 2^n(n + 1)

(method 2)

The simplicity of the method 1 is actually the result of iteration, let’s present it more explicitly.

x_{n + 2} = 4(x_{n + 1} - x_{n})
x_{n + 2} - 2x_{n + 1} = 2(x_{n + 1} - 2x_{n}) – – – – – – (*)

Now apply (*) again and again,

x_{n + 2} - 2x_{n + 1}
= 2(x_{n + 1} - 2x_{n}) [by (*)]
= 2^2(x_{n} - 2x_{n - 1}) [by (*)]
= 2^3(x_{n - 1} - 2x_{n - 2}) [by (*)]
. . . inductively . . .
= 2^n(x_{2} - 2x_{1}) [by (*)]

Thus,

x_{n + 2} = 2x_{n + 1} + 2^{n + 2} – – – – – – (#)
(\because x_1 = 4, x_2 = 12)

Now we apply (#) again and again,

x_{n + 2}
= 2x_{n + 1} + 2^{n + 2} [by (#)]
= 2(2x_n + 2^{n + 1}) + 2^{n + 2} [by (#)]
= 2^2x_n + 2(2^{n + 2})
= 2^2(2x_{n - 1} + 2^{n}) + 2(2^{n + 2}) [by (#)]
= 2^3x_{n - 1} + 3(2^{n + 2})
. . . inductively. . .
= 2^{n + 1}x_1 + (n + 1)2^{n + 2}
= 2^{n + 3} + (n + 1)2^{n + 2} (\because x_1 = 4)

Hence

x_{n + 2} = 2^{n + 3} + (n + 1)2^{n + 2}
x_{n} = 2^{n + 1} + (n - 1)2^{n} (replace n by n - 2)

Therefore

x_n = 2^n(n + 1)

2 則迴響 »

  1. 有少許簡單複雜化的感覺…
    不過我諗Nelson Sir的原意是想考我們能否看到要轉subject去做assumption而已…
    結果一條F.4 standard的數就考起了很多人…(包括我=.=")

    迴響 由 Patrick Wong — 2009/01/02 @ 11:36 下午 | 回覆

  2. Patrick, 是簡單複雜化了,用 M.I. 當然可以啦,但我其實是想,出題者究竟是如何擬出這題?

    迴響 由 johnmayhk — 2009/01/03 @ 9:07 下午 | 回覆


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