# Quod Erat Demonstrandum

## 2017/11/14

### as gs

Filed under: Additional / Applied Mathematics,mathematics,NSS — johnmayhk @ 12:29 上午
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Derive the formula for the sum of first $n$ terms of the following sequence in terms of $a,b,d,r,n$, where $r \ne 1$.

$ab,(a+d)br,(a+2d)br^2,(a+3d)br^3,\dots$

M1 同學立即得出 $\frac{1}{1/3}=3$

P(1 次) = $\frac{1}{3}$
P(2 次) = $\frac{2}{3}\frac{1}{3}$
P(3 次) = $\frac{2}{3}\frac{2}{3}\frac{1}{3}=(\frac{2}{3})^2\frac{1}{3}$
P(4 次) = $\frac{2}{3}\frac{2}{3}\frac{2}{3}\frac{1}{3}=(\frac{2}{3})^3\frac{1}{3}$
….

$E=1\times \frac{1}{3}+2\times (\frac{2}{3})\frac{1}{3} + 3\times (\frac{2}{3})^2\frac{1}{3} + 4\times (\frac{2}{3})^3\frac{1}{3} + \dots$

$E=p+2qp+3q^2p+4q^3p+\dots=p(1+2q+3q^2+4q^3+\dots)$

$S=1+2q+3q^2+4q^3+\dots$ ………………….. (1)

$qS=q+2q^2+3q^3+4q^4+\dots$ ………………….. (2)

(1) – (2) 得

$(1-q)S=1+q+q^2+q^3+...=\frac{1}{1-q}$

$E=pS=(1-q)S=\frac{1}{1-q}=\frac{1}{p}=\frac{1}{1/3}=3$

$S=1+2q+3q^2+4q^3+\dots$

$q+q^2+q^3+q^4+\dots = \frac{q}{1-q}$

$\Rightarrow \frac{d}{dq}(q+q^2+q^3+q^4+\dots )= \frac{d}{dq}\frac{q}{1-q}$

$\Rightarrow 1+2q+3q^2+4q^3+\dots = \frac{(1-q)-q(-q)}{(1-q)^2}=\frac{1}{(1-q)^2}$

$E$

$=ab+(a+d)br+(a+2d)br^2+(a+3d)br^3+\dots +(a+(n-1)d)br^{n-1}$

$=ab(1+r+r^2+\dots +r^{n-1})+dbr(1+2r+3r^2+\dots+(n-1)r^{n-2})$

$S=1+2r+3r^2+\dots+(n-1)r^{n-2}$

$S=\frac{1-nr^{n-1}+(n-1)r^n}{(1-r)^2}$

$r+r^2+r^3+\dots+r^{n-1}=\frac{r(1-r^{n-1})}{1-r}$

$\Rightarrow \frac{d}{dr}(r+r^2+r^3+\dots+r^{n-1}) = \frac{d}{dr}(\frac{r(1-r^{n-1})}{1-r})$

$\Rightarrow 1+2r+3r^2+\dots+(n-1)r^{n-2} = \frac{1-nr^{n-1}+(n-1)r^n}{(1-r)^2}$

$S=\frac{1-nr^{n-1}+(n-1)r^n}{(1-r)^2}$

（圖：授課員在學校生活（好似係））

「設某年頭存入銀行本金 $a$ 元，年利率 $r$%，複利息以年計，且每年年頭存入 $c$ 元。」

Let {$a_n$} be a sequence defined as

$a_n=\left \{ \begin{array}{ll} a \hspace{24 mm} \mbox{for}\hspace{5 mm} n = 0\\ba_{n-1}+c \hspace{10 mm} \mbox{for}\hspace{5 mm} n \ge 1\end{array}\right.$

where $b = 1+r$% $\ne 1$.

Derive the formula of $a_n$ in terms of $a,b,c,n$.

（這種題型其實在 applied mathematics 的概率 recurrence relation 一課也頗常見。）解法見下：

$a_n-k=b(a_{n-1}-k)$

$a_n=ba_{n-1}+(1-b)k$

$(1-b)k=c$

$k=\frac{c}{1-b}$

$a_n=ba_{n-1}+c \Rightarrow (a_n-\frac{c}{1-b})=b(a_{n-1}-\frac{c}{1-b})$

$d_n=a_n-\frac{c}{1-b}$

$d_n=bd_{n-1}$

$d_n=b^n(a-\frac{c}{1-b})$

$a_n=b^n(a-\frac{c}{1-b}) + \frac{c}{1-b}=ab^n+\frac{c(b^n-1)}{b-1}$

Let $a_1,a_2,a_3,a_4$ be non-zero real numbers such that

$a_2^4+a_3^4+a_1^2a_3^2+a_2^2a_4^2=2a_2a_3(a_1a_2+a_3a_4)$.

Prove that $a_1,a_2,a_3,a_4$ is a geometric sequence.

## 4 則迴響 »

1. 1 mark problem deserves a 1-line solution.
(a_1a_3 – a_2^2)^2 + (a_2a_4 – a_3)^2 = 0.

Hence, each term in the beacket is zero, and we are done.

迴響 由 apook — 2017/11/15 @ 4:23 下午 | 回應

2. 最後那 bonus 題，右邊好像有點不妥，是我做錯了嗎？
感覺應該是 2*a_2*a_3*(a_1*a_2+a_3*a_4)

迴響 由 Yutaka Juuichi — 2017/11/15 @ 5:32 下午 | 回應