Difficult Maths!!!! (prove推理)

2012-05-30 4:56 am

frequency:1,1,2,2,1,1,2,2,3,3,3,2,2,1,1,2,2,3,3,3,4,4,4,4............
(a)no. of term 2012,
(b)sum of first 100 terms
(c)first 20120913 terms,times of 9,12,13,20 have been appeared (total)

(option:term 10th:1,2,2,3,3,3,4....9,9,9,1,0,1,0,1,0,1........)

更新1:

re 001: 數列是這樣看的:1,1221,122333221,1223334444333221,.............

回答 (1)

Kara

2012-05-31 10:49 pm

✔ 最佳答案

(a)

T(1)=11=total 2 no.s
T(2)=2211=total 4 no.s
T(3)=223332211=total 9 no.s
T(4)=2233344443332211=total 16 no.s
T(5)=22333344445555544443332211=total 25 no.s
...........................................................

Ignore T(1). The increasing amount of very new term=n+(n-1)

Therefore,
T(6)=total 36 no.s
T(7)=total 49 no.s
T(8)=total 64 no.s
T(9)=total 81 no.s
.............................................................

The total no. of T(n)=n^2

Therefore, the no. of term 2012=2012^2=4048144//

(b)S(100)
=2+(2^2+3^2+4^2+5^2+6^2+7^2+8^2+9^2+10^2+...100^2)
=2+338349
=338351//

(c)In the 20120913 rd term, 9,12,13,20 have all appeared twice.

Hence, total times of 9,12,13,20 have appeared
=9*2+12*2+13*2+20*2
=108//

參考: MYSELF AND EXCEL

👍 0 👎 0