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樂仔 2007-06-03 16:28

有條MI唔識做...求助

(n+1)^2+(n+2)^2+(9n+3)^2+...+(2n)^2 = [n(2n+1)(7n+1)]/6



幫幫手

BBF 2007-06-03 17:38
引用第0樓樂仔於2007-06-03 16:28發表的“有條MI唔識做...求助”:
(n+1)^2+(n+2)^2+9n+3)^2+...+(2n)^2 = [n(2n+1)(7n+8)]/6
幫幫手 [表情]


(n+1)^2+(n+2)^2+9n+3)^2+...+(2n)^2 = [n(2n+1)(7n+1)]/6

ryan_kewell_7 2007-06-03 17:41
引用第1樓ballballfull於2007-06-03 17:38發表的“”:
(n+1)^2+(n+2)^2+9n+3)^2+...+(2n)^2 = [n(2n+1)(7n+1)]/6
你咩都識既

BBF 2007-06-03 17:51
引用第2樓ryan_kewell_7於2007-06-03 17:41發表的“”:
你咩都識既 [表情]

樂仔 2007-06-03 17:54
引用第1樓ballballfull於2007-06-03 17:38發表的“”:
(n+1)^2+(n+2)^2+9n+3)^2+...+(2n)^2 = [n(2n+1)(7n+1)]/6

唔明

BBF 2007-06-03 18:05
引用第4樓樂仔於2007-06-03 17:54發表的“”:
[表情] [表情] 唔明


題目錯左...

(n+1)^2+(n+2)^2+9n+3)^2+...+(2n)^2 = [n(2n+1)(7n+1)]/6

n=1
LHS=(1+1)^2=4
RHS=1x3x8/6=4
P(1) is true

when n=k+1
LHS= (k+2)^2 + (k+3)^2 + ... + (2k)^2 + (2K+1)^2 + (2k+2)
= {[k(2k+1)(7k+1)]/6 - (k+1)^2} + (2K+1)^2 + (2k+2)
= (14k^3+51K^2+61K+24)/6

RHS= [(k+1)(2k+3)(7k+8)]/6
= (14k^3+51K^2+61K+24)/6

LHS = RHS
i,e. P(k+1) is true, ......

(我都唔記得可唔可以２邊一齊做-.-", 如果得就做完)

垃圾桶 2007-06-03 18:16
引用第5樓ballballfull於2007-06-03 18:05發表的“”:
(我都唔記得可唔可以２邊一齊做-.-", 如果得就做完)


唔得咪掉轉來抄第二part

BBF 2007-06-03 18:21
引用第6樓垃圾桶於2007-06-03 18:16發表的“”:
唔得咪掉轉來抄第二part


ｎ＋１　年前ｏ既野，．．唔記得．．．
何況我ａｍａｔｈ只是得ｃ．．．

如果唔得，老屈佢＝果ｐａｔ野
事實佢地係相等

樂仔 2007-06-03 18:21

唔該哂

樂仔 2007-06-03 18:36
引用第5樓ballballfull於2007-06-03 18:05發表的“”:
題目錯左...
(n+1)^2+(n+2)^2+9n+3)^2+...+(2n)^2 = [n(2n+1)(7n+1)]/6
.......

呢步點解呀?
LHS= (k+2)^2 + (k+3)^2 + ... + (2k)^2 + (2K+1)^2 + (2k+2)

BBF 2007-06-03 18:57
引用第9樓樂仔於2007-06-03 18:36發表的“”:
呢步點解呀?
LHS= (k+2)^2 + (k+3)^2 + ... + (2k)^2 + (2K+1)^2 + (2k+2)


(n+1)^2+(n+2)^2+(n+3)^2+...+(2n)^2

put n = k+1

(k+1+1)^2+(k+1+2)^2+(k+1+3)^2+...+[2(k+1)]^2
由於係+1,+1,+1咁上, 直至自己的2倍
2(K+1)之前係有 2K+1

即由本來
1,2,3,4,...,(2n-1), 2n
=>
2,3,4,5,...,(2k-1), 2k, 2k+1, 2k+2

ie, 多左 第"2k+1", 第"2k+2", 少左 第"1"

樂仔 2007-06-03 19:02
引用第10樓ballballfull於2007-06-03 18:57發表的“”:
(n+1)^2+(n+2)^2+(n+3)^2+...+(2n)^2
put n = k+1
.......

呢個我明...但係點解最尾果個唔駛square既?

BBF 2007-06-03 19:08
引用第11樓樂仔於2007-06-03 19:02發表的“”:
呢個我明...但係點解最尾果個唔駛square既?


打漏左,....死未

我寫完先打,...打漏左,...
好彩唔係考試啫

樂仔 2007-06-03 19:19
點解我會計到咁既?
(k+2)^2+(k+3)^2+(k+4)^2+...+(2k)^2+(2k+2)^2

中間少左個(2k+1)^2既????

同埋點出黎ga? 乜唔係代n=k+1咩?

樂仔 2007-06-03 19:20
= {[k(2k+1)(7k+1)]/6 - (k+1)^2} + (2K+1)^2 + (2k+2)

點解要減(k+1)^2?

即係將上一步冇寫到既(k+1)^2搬返落黎咁解?

BBF 2007-06-03 19:27
引用第13樓樂仔於2007-06-03 19:19發表的“”:
點解我會計到咁既?
(k+2)^2+(k+3)^2+(k+4)^2+...+(2k)^2+(2k+2)^2
中間少左個(2k+1)^2既????
.......


你要明左條式係點行先
佢係由1開始, 1下+1, 1,2,3,4,5,...,n-2, n-1, n

條式最後係 (2n)^2
當put左k+1入去
即最後係 [2(k+1)]^2
i.e, (2k+2)^2
由於條式係+1,+1咁上
咁(2k+2)^2之前係有(2k+1)^2, (2k)^2, (2k-1)^2,....

BBF 2007-06-03 19:30
引用第14樓樂仔於2007-06-03 19:20發表的“”:
= {[k(2k+1)(7k+1)]/6 - (k+1)^2} + (2K+1)^2 + (2k+2)
點解要減(k+1)^2? [表情]
即係將上一步冇寫到既(k+1)^2搬返落黎咁解?


(n+1)^2+(n+2)^2+(9n+3)^2+...+(2n)^2 = [n(2n+1)(7n+1)]/6

因為put左n=k+1, 就冇左第一舊

(n+1)^2 + (n+2)^2+(9n+3)^2+...+(2n)^2 = [n(2n+1)(7n+1)]/6

(n+2)^2+(9n+3)^2+...+(2n)^2 = [n(2n+1)(7n+1)]/6 - (n+1)^2

(n+2)^2+(9n+3)^2+...+(2n)^2 + [(2n+1)^2 + (2n+2)^2] = [n(2n+1)(7n+1)]/6 - (n+1)^2 + [(2n+1)^2 + (2n+2)^2]

樂仔 2007-06-03 19:34
引用第16樓ballballfull於2007-06-03 19:30發表的“”:
(n+1)^2+(n+2)^2+(9n+3)^2+...+(2n)^2 = [n(2n+1)(7n+1)]/6
因為put左n=k+1, 就冇左第一舊
.......

明白

樂仔 2007-06-03 19:36
哦...明哂啦!


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