這條M.I.怎樣做？

好似我做過....

寫曬出黎先

Let P(n) be the proposition
" 1/(2x5) + 1/(5x8) + 1/(8+11) + ... + 1/(3n-1)(3n+2) = n/(6n+4) "

When n = 1,
L.H.S. = 1/(2x5) = 1/10
R.H.S. = 1/(6+4) = 1/10
so L.H.S. = R.H.S.
so P(1) is true

Assume that P(k) is true for some positive integers k
i.e. 1/(2x5) + 1/(5x8) + 1/(8+11) + ... + 1/(3k-1)(3k+2) = k/(6k+4)

When n = k+1
L.H.S. = 1/(2x5) + 1/(5x8) + 1/(8+11) + ... + 1/(3k-1)(3k+2) + 1/[3(k+1)-1][3(k+1)+2]
= k/(6k+4) + 1/[3k+3-1][3k+3+2]
= k/2(3k+2) + 1/(3k+2)(3k+5) ........................抽commom factor
= k(3k+5)/2(3k+2)(3k+5)　+　2/2(3k+2)(3k+5)
= [k(3k+5) + 2]/2(3k+2)(3k+5) ................+-數
= (3k^2 + 5k + 2) / 2(3k+2)(3k+5) ................expan
= (3k + 2)(k + 1) / 2(3k+2)(3k+5) .........factorazition
= (k + 1) / 2(3k+5) ...................約
= (k + 1) / (6k + 10)

R.H.S. = (k+1)/[6(k+1)+4]
= (k+1) / [6k+6+4]
= (k+1) / (6k + 10)

so L.H.S. = R.H.S.
SO P(k+1) is true

By the principle of mathematical induction , P(n)is true for all positive integers n



don't copy me

2007-11-26 18:51:02 補充：
人地出得呢d數......基本上條條都一定岩.......我老師話只要背曬d格式...會考個題都會有一半分

2007-11-26 18:51:38 補充：
只要背曬d格式...唔識做都好..寫d格式....會考個題都會有一半分

其實你唔使計都會知道個答案係咩...
因為用右方既轉做(k+1)就係...
你計o個陣第一時間抽常數,(k+1)出lai
再抽賣其他咁就做完嫁喇...
MI其實條條都係用同一個模式做...

When n = k+1,
LHS = 1/(2x5) + 1/(5x8) + ... + 1/(3k-1)(3k+2) + 1/(3k+2)(3k+5)
= k/(6k+4) + 1/(3k+2)(3k+5)
= k/2(3k+2) + 1/(3k+2)(3k+5)
= [k(3k+5) + 2]/2(3k+2)(3k+5)
= (3k^2 + 5k + 2)/2(3k+2)(3k+5)
= (3k+2)(k+1)/2(3k+2)(3k+5)
= (k+1)/(6k+10)
= (k+1)/[6(k+1)+4]
The statement is true for n = k+1.