遞迴數列問題 如何化簡成一般項?

2011-05-17 9:24 pm
3/4 X 8/9 X 15/16 X.......X (n^2 -1)/ n^2 =??


請詳解過程

答案為n+1/2n

回答 (5)

2011-05-17 10:08 pm
✔ 最佳答案
(n^2-1)/n=(n-1)(n+1)/n^2分子=(1*3)(2*4)(3*5)(4*6)…[(n-2)*n][(n-1)*(n+1)=1*2*3^2*4^2*5^2*…*(n-2)^2*(n-1)^2*n*(n+1)分母=2^2*3&2*4^2*…*n^2分子分母消去: 3^2*4^2*…*(n-1)^2原式=1*2*n(n+1)/[2^2*n^2]=(n+1)/2n[[Done]]用歸納法也很容易正得此結果。

2011-05-17 21:02:46 補充:
To :Sam ( 實習生 3 級 )
太好玩了!真是有緣,空中乾一杯。
2011-05-18 8:21 am
"月下"的回答簡捷!
2011-05-17 10:22 pm
(3/4)*(8/9)*(15/16)*….*[(n^2-1)/n^2]=??
請詳解過程
Sol
(3/4)*(8/9)*(15/16)*….*[(n^2-1)/n^2]
=(1*3/4)*(2*4/9)*(3*5/16)*….*[(n-1)(n+1)/n^2]
=(1*3/4)*(2*4/9)*(3*5/16)*(4*6/25)*(5*7/36)*….*[(n-5)*(n-3)/(n-4)^2]
*[(n-4)*(n-2)/(n-3)^2]*[(n-3)*(n-1)/(n-2)^2]*[(n-2)*n/(n-1)^2]
*[(n-1)(n+1)/n^2]
=(1/4)*(2/3)*(3/4)*(4/5)*(5/6)*….*[(n-5)/(n-4)]*[(n-4)/(n-3)]
*[(n-3)/(n-2)]*[(n-2)/(n-1)]*[(n-1)(n+1)/n]
=(1/4)*(2/1)*(1/1)*(1/1)*(1/1)*….*(1/1)*(1/1)*(1/1)*(1/1)*[1*(n+1)/n]
=(n+1)/(2n)


2011-05-17 10:20 pm
(n^2 -1)/ n^2=[(n -1)/ n][(n+1)/n]
n=2~n
(1/2) {(3/2)(2/3) (4/3)(3/4) (5/4)(4/5) (6/5)…. [(n -1)/ n] } [(n+1)/n]
=(1/2)( n+1)/n=(n+1)/(2n)

中間都會消掉,最後剩下頭尾兩項。

2011-05-17 14:21:45 補充:
我也奇怪,怎麼能回答兩次?
2011-05-17 10:12 pm
您好
先從最後項著手,項數之間的關係為連乘,n從2開始把(n^2-1)/(n^2)化簡為1-(1/n)^2=[1+(1/n)][1-(1/n)]先看[1+(1/n)]的規律(3/2)*(4/3)*...*[n/(n-1)]*[(n+1)/n)]=(n+1)/2...(一)再看[1-(1/n)]的規律(1/2)*(2/3)*...*[(n-2)/(n-1)]*[(n-1)/n)]=1/n...(二)最後把(一)(二)相乘即可得證(n+1)/(2n)

2011-05-17 14:13:31 補充:
我看到其他發問者嚇到了,我想我怎會回答兩次XD
(另外一個Sam大大,安安)

2011-05-18 09:34:50 補充:
TO:Sam ( 初學者 1 級 ) CHEERS!
參考: 自己的淺見


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