費費 4-14

[u9069700]

4-14
n is the sum of the first K consecutive positive integers, where 101<=k<=150. what is the sum of 1/n?
answer:2(1/101-1/151)
這一題我是醬算的:n=[(1+k)k]/2 .... 1/n=2/[(1+k)k]
1/[k(k+1)]=1/k-1/(k+1)
又101<=k<=150
則1/101>=1/k>=1/150
阿接下就不會了?因為不等式的運算法忘光了?.....><

[huacy]

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[u9069700]

ㄚ阿....抱歉抱歉
我PO之前有搜尋過但沒找到
可能是不夠仔細八
去好好研究一下搜尋功能好了!!!
謝謝!!!!