# Quod Erat Demonstrandum

## 2008/11/18

### Create an m.i. question

It is not difficult to create questions like:

Prove by mathematical induction that

$\frac{3^3\times1}{4!} + \frac{3^4\times2}{5!} + \frac{3^5\times3}{6!} + \dots + \frac{3^{n+2}\times n}{(n+3)!} = \frac{9}{2} - \frac{3^{n+3}}{(n+3)!}$

for any positive integer $n$.

It is all about the so-called telescoping principle.

What’s it? It’s just one of the methods of finding the sum (or product) of terms.

To evaluate the sum of

$a_1 + a_2 + a_3 + \dots + a_n$ (say),

it will be easy when we can split each term into

$a_k = b_k - b_{k+1}$.

See

$a_1 + a_2 + a_3 + \dots + a_n$
= $(b_1 - b_2) + (b_2 - b_3) + (b_3 - b_4) + \dots + (b_n - b_{n+1})$
= $b_1 - b_{n+1}$

Thus, we may create an m.i. question in the form of

$a_1 + a_2 + a_3 + \dots + a_n = b_1 - b_{n+1}$"

To set up a common M.I. question, just take ‘reasonable’ $b_k$.

For example, we may take

$b_k = \frac{a^k}{k!}$

then

$b_k - b_{k+1} = \frac{a^k}{k!} - \frac{a^{k+1}}{(k+1)!} = \frac{a^k(k + 1 - a)}{(k+1)!}$

For the so-called beauty of the question, we may start from k = a, that is

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

Sum up the above $n$ equations, we have

$\frac{a^a}{a!} - \frac{a^{a+n}}{(a+n)!} = \frac{a^a(1)}{(a+1)!} + \frac{a^{a+1}(2)}{(a+2)!} + \frac{a^{a+2}(3)}{(a+3)!} + \dots +\frac{a^{a+2}(n)}{(a+n)!}$ – – – – – – (*)

By putting $a = 1, 2, 3, 4, \dots$ into (*), we can create the following m.i. questions for students revision:

Prove the following are true for any positive integer $n$,

1. $\frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!} + \dots +\frac{n}{(n+1)!} = 1 - \frac{1}{(n+1)!}$
2. $\frac{2^2(1)}{3!} + \frac{2^3(2)}{4!} + \frac{2^4(3)}{5!} + \dots +\frac{2^{n+1}(n)}{(n+2)!} = 2 - \frac{2^{n+2}}{(n+2)!}$
3. $\frac{3^3(1)}{4!} + \frac{3^4(2)}{5!} + \frac{3^5(3)}{6!} + \dots +\frac{3^{n+2}(n)}{(n+3)!} = \frac{9}{2} - \frac{3^{n+3}}{(n+3)!}$ (The original question above)
4. $\frac{4^4(1)}{5!} + \frac{4^5(2)}{6!} + \frac{4^6(3)}{7!} + \dots +\frac{4^{n+3}(n)}{(n+4)!} = \frac{32}{3} - \frac{4^{n+4}}{(n+4)!}$
$\dots$

Furthermore, if we take

$b_k = \sin(k)$, then

$b_{k+1} - b_k = \sin(k+1) - \sin(k) = 2\cos\frac{2k+1}{2}\sin\frac{1}{2}$

Hence, we have

$\sin(2) - \sin(1) = 2\cos\frac{3}{2}\sin\frac{1}{2}$
$\sin(3) - \sin(2) = 2\cos\frac{5}{2}\sin\frac{1}{2}$
$\sin(4) - \sin(3) = 2\cos\frac{7}{2}\sin\frac{1}{2}$
$\sin(5) - \sin(4) = 2\cos\frac{9}{2}\sin\frac{1}{2}$
$\dots$
$\sin(n) - \sin(n - 1) = 2\cos\frac{2n-1}{2}\sin\frac{1}{2}$

sum up the above, yields

$\sin(n) - \sin(1) = 2\sin\frac{1}{2}(\cos\frac{3}{2} + \cos\frac{5}{2} + \cos\frac{7}{2} + \cos\frac{9}{2} + \dots + \cos\frac{2n-1}{2})$

Thus, we create a question like:

prove by mathematical induction that

$\cos\frac{3}{2} + \cos\frac{5}{2} + \cos\frac{7}{2} + \cos\frac{9}{2} + \dots + \cos\frac{2n-1}{2} = \frac{\sin(n) - \sin(1)}{2\sin\frac{1}{2}}$

for all integers $n \ge 2$.

Students, try to let other $b_k$ to create some boring m.i. questions at your command.

## 7 則迴響 »

1. John Sir

好耐之前係濟濟一堂… 你地 post 過一個由 David Morin 教授寫既一本 ebook, 內容係關於 mechanics and special relativity….
唔知你有無 save 低舊既 pdf file…
有既可以 send 俾我嗎？

thx

迴響 由 Ricky — 2008/11/18 @ 11:18 下午 | 回應

2. AL時無留心上堂,或自己是無天份,總是學不懂telescoping principle…
不過還是感謝阿sir的教導…

迴響 由 lanven — 2008/11/19 @ 4:32 上午 | 回應

3. @Ricky

對不起，我沒有存底。該網頁似乎有一些 lecture notes

http://www.courses.fas.harvard.edu/~phys16/2004_lectures/

不知可以用得著。

@lanven

節節相消原理本身很簡單，但有時應用起來可以是很難的，不一定是天份問題。謝謝大家留言！

迴響 由 johnmayhk — 2008/11/19 @ 11:26 上午 | 回應

4. 在我的印象中, 所有課本的等比級數都是節節相消的變身.

迴響 由 chin — 2012/05/14 @ 1:48 上午 | 回應

5. 在我的印象中, 所有課本的等比級數都是節節相消的變身.

迴響 由 mm100100 — 2012/05/14 @ 1:52 上午 | 回應

6. Hi, the last term on the LHS of the trigo-related induction question should be cos {(2n – 1)/2} and that we prove for n >= 2. Thanks for sharing!

迴響 由 wenshih — 2012/05/19 @ 12:23 下午 | 回應

• Corrected. Thank you Wenshih! Long time no see!

迴響 由 johnmayhk — 2012/05/21 @ 6:30 上午 | 回應