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一個人的狂歡

 
 
 

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Just for fun  

2015-08-19 23:44:39|  分类: 千色塵世 |  标签: |举报 |字号 订阅

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有人在微信求助。我没兴趣回答这个人。不过我觉得做做这些数学题也挺有趣的。题目已经给了提示了:“裂项”。
Just for fun - Dreaming -               一個人的狂歡
现在很晚了,我就做了两题。
1)原式 = 1 + (1/1)*(1/3)+ (1/2)*(1/3) + (1/2)*(1/5) + (1/3)*(1/5) + (1/3)*(1/7) + (1/4)*(1/7)
= 1 + (1/3)*(1+1/2) + (1/5)*(1/2+1/3) + (1/7)*(1/3+1/4)
= 1 + 1/2 + 1/6 + 1/12
= (12 + 6 + 2 +1) / 12
= 21/12 = 7/4

2)原式 = (1/2 + 2/1) + (2/3 + 3/2) + (3/4 + 4/3) + ... + (2002/2003 + 2003/2002)
= (0/1 + 2/1) + (1/2 + 3/2) + (2/3 + 4/3) + (3/4 + 5/4) + ... + [(k-1)/k + (k+1)/k] + ... + (2001/2002 + 2003/2002) + 2002/2003
= 2*2002 + 2002/2003

第二天。
3)原式 = (1 + 1/(1*3)) + (1 + 1/(3*5)) + (1 + 1/(5*7)) + ... + (1 + 1/(13*15))
= 7 + [1/(1*3) + 1/(3*5) + 1/(5*7) + ... + 1/(13*15)]
= 7 + [(1/1 - 1/3)*(1/2) + (1/3 - 1/5)*(1/2) + (1/5 - 1/7)*(1/2) + ... + (1/13 - 1/15)*(1/2)]
= 7 + (1/2) * (1/1 - 1/3 + 1/3 - 1/5 +1/5 - 1/7 + ... + 1/11 - 1/13 + 1/13 - 1/15)
= 7 + (1/2) * (1/1 - 1/3 + 1/3 - 1/5 +1/5 - 1/7 + ... + 1/11 - 1/13 + 1/13 - 1/15)
= 7 + (1/2) * (1/1 - 1/15)
= 7 + 7/15

4)原式 = (20 + 21) / 60 
= 1/3 + 7/20
= 1/3 + 1/4 + 1/5 - 1/10
= 1/3 + 1/4 + 1/10

5)原式 = (1+2) / [(1*2)^2] + (2+3) / [(2*3)^2] + ... + [k + (k+1)] / { [k * (k+1)]^2 }  + ... + (7+8) / [(7*8)^2]
= 1 / (1^2) - 1 / (2^2) + 1 / (2^2) - 1 / (3^2) + ... + 1 / (k^2) - 1 / [(k+1)^2] + ... + 1 / (6^2) - 1 / (7^2) + 1 / (7^2) - 1 / (8^2)
= 1 / (1^2) - 1 / (2^2) + 1 / (2^2) - 1 / (3^2) + ... + 1 / (k^2) - 1 / [(k+1)^2] + ... + 1 / (6^2) - 1 / (7^2) + 1 / (7^2) - 1 / (8^2)
= 1 - 1/64
= 63/64

6)原式 = 1/3 * [1/1 + 1/(1+2) + 1/(1+2+3) + ... + 1/(1+2+3+...+30)]
= 1/3 * { 2/[(1+1)*1] + 2/[(2+1)*2] + 2/[(3+1)*3] + ... + 2/[(k+1)*k] + ... +2/[(30+1)*30] }
= 1/3 * { (1/1 - 1/2)*2 + (1/2 - 1/3)*2 + (1/3 - 1/4)*2 + ... + [1/k - 1/(k+1)]*2 + ... + (1/30 - 1/31)*2 }
= 2/3 * [ 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/k - 1/(k+1) + ... + 1/29 - 1/30 + 1/30 - 1/31 ]
= 2/3 * [ 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/k - 1/(k+1) + ... + 1/29 - 1/30 + 1/30 - 1/31 ]
= 2/3 * (1 - 1/31)
= 20/31

7)原式 = (2 - 1) / 2! + (3 - 1) / 3! + (4 -1) / 4! + ... + (k - 1) / k! + ... + (10 - 1) / 10!
= 1/1! - 1/2! +1/2! - 1/3! + 1/3! - 1/4! + ... + 1/(k - 1)! - 1/k! + ... + 1/8! - 1/9! + 1/9! - 1/10!
= 1/1! - 1/2! +1/2! - 1/3! + 1/3! - 1/4! + ... + 1/(k - 1)! - 1/k! + ... + 1/8! - 1/9! + 1/9! - 1/10!
= 1 - 1/10!

8)原式 = { [ (7*5) / (2*3) - (7*7) / (3*4) + (7*9) / (4*5) - (7*11) / (5*6) + (7*13) / (6*7) - (7*15) / (7*8) ] - 11/8 } * 8
= { 7 * [ 5/(2*3)  - 7/(3*4) + 9/(4*5) - 11/(5*6) + 13/(6*7) - 15/(7*8) ] - 11/8 } * 8
= { 7 * [ (1/2 + 1/3) - (1/3 +1/4) + (1/4 +1/5) - (1/5 + 1/6) + (1/6 + 1/7) - (1/7 + 1/8) ] - 11/8 } * 8
= { 7 * [ (1/2 + 1/3) - (1/3 +1/4) + (1/4 +1/5) - (1/5 + 1/6) + (1/6 + 1/7) - (1/7 + 1/8) ] - 11/8 } * 8
= [ 7 * (1/2 - 1/8) - 11/8 ] * 8
= 10

第9题卡住了。

10)原式 = 2010/2008 + (2010*2009) / (2008*2007) + (2010*2009) / (2007*2006) + (2010*2009) / (2006*2005) + ... + (2010*2009) / [k*(k-1)] + ... + (2010*2009) / (3*2) + (2010*2009) / (2*1)
= 2010/2008 + (2010*2009) * { (1/2007 - 1/2008) + (1/2006 - 1/2007) + (1/2005 - 1/2006) + ... + [1/(k-1) - 1/k] + ... + (1/2 - 1/3) + (1/1 - 1/2) }
= 2010/2008 + (2010*2009) * { (1/2007 - 1/2008) + (1/2006 - 1/2007) + (1/2005 - 1/2006) + ... + [1/(k-1) - 1/k] + ... + (1/2 - 1/3) + (1/1 - 1/2) }
= 2010/2008 + (2010*2009) * (1 - 1/2008)
= 2010/2008 + 2010*2009 - (2010*2009) / 2008
= 2010*2009 - 2010
= 2010 * 2008

第11题卡住了。

12)原式 = 1155 * { (3+2)/(2*3*4) + (4+3)/(3*4*5) + ... + (k+k-1) / [ (k-1)*k*(k+1) ] + ... + (9*8)/(8*9*10) + (10+9)/(9*10*11) }
= 1155 * { [1/(2*4) + 1/(3*4)] + [1/(3*5) + 1/(4*5)] + ... + 1/[(k-1)*(k+1)] + 1/[k*(k+1)] + ... + [1/(8*10) + 1/(9*10)] + [1/(9*11) + 1/(10*11)] }
= 1155 * { (1/2 - 1/4)*(1/2) + 1/3 - 1/4 + (1/3 - 1/5)*(1/2) + 1/4 - 1/5 + ... + [1/(k-1) - 1/(k+1)]*(1/2) + 1/k - 1/(k+1) + ... + (1/8 - 1/10)*(1/2) + 1/9 -1/10 + (1/9 - 1/11)*(1/2) + 1/10 - 1/11 }
= 1155 * { [1/3 - 1/4 + 1/4 - 1/5 + ... + 1/k - 1/(k+1) + ... + 1/9 -1/10 + 1/10 - 1/11] + (1/2)* [1/2 - 1/4 + 1/3 - 1/5 + 1/4 - 1/6 + 1/5 - 1/7... + 1/(k-1) - 1/(k+1) + ... + 1/8 - 1/10 + 1/9 - 1/11] }
= 1155 * { [1/3 - 1/4 + 1/4 - 1/5 + ... + 1/k - 1/(k+1) + ... + 1/9 -1/10 + 1/10 - 1/11] + (1/2)* [1/2 - 1/4 + 1/3 - 1/5 + 1/4 - 1/6 + 1/5 - 1/7... + 1/(k-1) - 1/(k+1) + ... 1/7 - 1/9 + 1/8 - 1/10 + 1/9 - 1/11] }
= 1155 * { (1/3 - 1/11) + [ (1/2) * (1/2 + 1/3 - 1/10 - 1/11) ] }
= 1155 * [ (1/2) * (1/2 - 1/10) + (1/2) * (1/3 - 1/11) + (1/3 - 1/11) ]
= 1155 * [1/5 + (3/2) * (1/3 - 1/11)]
= 1155 * [ (22 + 55 -15) / 110 ]
= 31 * 21

第三天
第9题看来只能硬解了,虽然解法不怎么简洁优雅,但能得到答案。
9)原式 = (1/3) * (1/23) * (1/29)
= (1/3 + 1/23) * (1/26) * (1/29)
(1/3 + 1/23) * (1/26 + 1/29) * (1/55)
= [ (1/3) * (1/26) + (1/3) * (1/29) + (1/23) * (1/26) + (1/23) * (1/29) ] * (1/55)
= (1/78 + 1/87 + 1/598 + 1/667) * (1/55)
= 1/4290 + 1/4785 + 1/32890 + 1/36685

我觉得第11题是里面最难的一题了。
11)构造一个数列:
A1 = (2/6) * (1 + 1/5) = 2/5
A2 = (3/7) * (1 + A1) = 3/5
A3 = (4/8) * (1 + A2) = 4/5
...
Ak = [(k+1) / (k +5)] * (1 + Ak-1) = (k+1) / 5
...
A44 = (45/49) * (1 + A43) = 45/5
A45 = (46/50) * (1 + A44) = 46/5
A46 = (47/51) * (1 + A45) = 47/5
原式 = 1/52 * [ 1 + 47/51 + (47*46) / (51*50)  + (47*46*45) / (51*50*49) + ... + (47*46*45*...*2*1) / (51*50*49*...*6*5) ]
= (1/52) * { 1+ 47/51 * [ 1 + 46/50 + (46*45) / (50*49) + ... + (46*45*...*2*1) / (50*49*...*6*5)] }
(1/52) * ( 1+ 47/51 * { 1 + 46/50 * [ 1 + 45/49 + ... + (45*...*2*1) / (49*...*6*5) ] } )
...
(1/52) * { 1+ 47/51 * [ 1 + 46/50 * ( 1 + 45/49 * { 1 + ... + 3/7 * [ 1+ (2/6) *(1 + 1/5) ] } ) ] }
(1/52) * [ 1+ 47/51 * ( 1 + 46/50 * { 1 + 45/49 * [ 1 + ... + 3/7 * ( 1+ A1 ) ] } ) ]
(1/52) * [ 1+ 47/51 * ( 1 + 46/50 * { 1 + 45/49 * [ 1 + ... + 4/8 * ( 1 + A2 ) ] } ) ]
...
(1/52) * { 1+ 47/51 * [ 1 + 46/50 * ( 1 + A44 ) ] }
(1/52) * [ 1+ 47/51 * ( 1 + A45 ) ) ]
(1/52) * (1+ A46)
= (1/52) * [1+ (47/5) ]
= 1/5

总结:其实不外乎就是两种技巧。
1) 1 / (a*b) = (1/a + 1/b) * 1 / (a + b),或1 / (a*b) = (1/a - 1/b) * 1 / (b - a)。后一种往往更有用,因为有减法,可能可以各项互相消去。
2)如果是一个数列,则观察这个数列的通项公式,寻找1 / (a*b) 这种形式。
这样的练习,做到一定量之后,其实就是在被训练成解题机器而已,没卵用。
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