數學思考題

2009-10-15 7:03 am
1. (1/1_1/2+1/3+1/4…+1/100)+(2/2+2/3+2/4+…+2/100)+(3/3+3/4+3/5+…3/100)+…+(99/99+99/100)+(100/100)

2. 2^1997+3^1998+4^1999+5^2000+6^2001+7^2002的個位數字是什麼?


有沒有人可以幫小弟計這兩條數呢...

請列舉步驟 謝
更新1:

四個一組1997 = 499*4 + 1所以2^1997個位為2 ..? 唔明?點解499*4 + 1 之後2^1997個位為2 499*4 +1同2^1997個位為2有咩關係?

回答 (2)

2009-10-15 7:47 am
✔ 最佳答案
1. S = (1/1+1/2+1/3+1/4…+1/100)+(2/2+2/3+2/4+…+2/100)+(3/3+3/4+3/5+…3/100)+…+(99/99+99/100)+(100/100)
= (1/1) + (1/2 + 2/2) + (1/3 + 2/3 + 3/3) + (1/4 + 2/4 + 3/4 + 4/4) + (1/5 + 2/5 + 3/5 + 4/5 + 5/5) + .... + (1/99 + 2/99 + ... + 99/99) + (1/100 + 2/100 + ... + 100/100)
通項 = 1/n + 2/n + 3/n + ... + n/n)
= (1/n)(1 + 2 + ... + n)
= (1/n)(n)(n + 1)/2
= (n + 1)/2
S = (1 + 1)/2 + (2 + 1)/2 + (3 + 1)/2 + ... + (100 + 1)/2
= (1/2)(1 + 2 + 3 + ... + 100) + (1/2)(100)
= (1/2)(100)(101)/2 + 50
= 2525 + 50
= 2575
2. 2^1997+3^1998+4^1999+5^2000+6^2001+7^2002的個位數字是什麼?
2的次方為2, 4, 8, 16,32,64, 128, 256... 個位順2,4,8,6,2,4,8,6...進行
四個一組1997 = 499*4 + 1所以2^1997個位為2
3的次方為3,9,27,81,243,729,2187,6561...個位順3,9,7,1...進行
四個一組1998 = 499*4 + 2所以3^1998個位為9
4的次方為4,16,64,256是4,6,4,6進行
1999 = 2*999 + 1; 4^1999 個位為4
5的次方個位必為5
6的次方個位必為6
7的次方為7,49,343,2401,16807...順7,9,3,1進行
2002 = 4*500 + 2; 7^2002個位為9
2^1997+3^1998+4^1999+5^2000+6^2001+7^2002的個位數字要看
2 + 9 + 4 + 5 + 6 + 9 (=35)的個位 = 5
2009-10-15 7:37 am
Set A=(1/1+1/2+1/3+1/4…+1/100)+(2/2+2/3+2/4+…+2/100)
+(3/3+3/4+3/5+…3/100)+ …+(99/99+99/100)+(100/100)
=(1/1)+(1/2+2/2)+(1/3+2/3+3/3)+(1/4+2/4+3/4+4/4)+...+(1/100+2/100+...+100/100)

2009-10-14 23:38:05 補充:
=(1/1)*(1)+(1/2)*(1+2)+(1/3)*(1+2+3)+...+(1/100)*(1+2+3+...+100)
2A=2*(1/1)*(1)+2*(1/2)*(1+2)+2*(1/3)*(1+2+3)+...+2*(1/100)*(1+2+3+...+100)
=2+(1/2)*(2*3)+(1/3)*(3*4)+...+(1/100)*(100*101)
=2+3+4+...+101
=(101*102)/2-1
=5150
A=2575


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