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L K

2009-11-12 5:58 am

1∕1 + 1∕1+2 + 1∕1+2+3 + 1∕1+2+3+4 + 1∕1+2+3+4+5 + ... 1∕1+2+3+...+100 = ?

回答 (3)

用戶2053

2009-11-12 6:12 am

✔ 最佳答案

1/1 + 1/(1+2) + 1/(1+2+3) + 1/(1+2+3+4) +...+ 1/(1+2+3+...+100)

= 1/[1*(1+1)/2] + 1/[2(1+2)/2] + 1/[3(1+3)/2] + 1/[4(1+4)/2 +...+ 1/[100(1+100)/2]

= 2/(1*2) + 2/(2*3) + 2/(3*4) + 2/(4*5) +...+ 2/(100*101)

= 2(1 - 1/2) + 2(1/2 - 1/3) + 2(1/3 - 1/4) + 2(1/4 - 1/5) + ... + 2(1/100 - 1/101)

= 2(1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - ... - 1/100 + 1/100 - 1/101)

= 2(1 - 1/101)

= 2(100/101)

= 200/101

公式 : 1+2+3+...+N= N(1+N)/2

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