推 christianSK:聖誕快樂~ 12/25 22:49
→ aoqq12:xd 還在想就有解答了 g聖誕快樂 12/25 22:51
推 compulsory:謝謝 剛剛做到這題完全沒有頭緒 12/25 22:58
推 qk211:那個 (4+(n-1)/n)(a1+a2+a3+.....+an-1) 應該是這樣吧 12/26 23:41
※ 引述《compulsory (生既無歡 死又何懼?)》之銘言：
: http://120.126.115.57/library/download/collection/exam/graduate/dci/dci993.pdf
: 第三題
: an=(4/n)(a1+a2+a3+.....+an-1) a1=1
: 該怎麼解?
an=(4/n)(a1+a2+a3+.....+an-1) a1=1
兩邊同+(a1+a2+a3+.....+an-1)
a1+a2+a3+.....+an-1+an=(4+n/n)(a1+a2+a3+.....+an-1)
令 Sn = a1+a2+a3+.....+an
=> Sn = (4+n/n)Sn-1 = ... = (n+4)(n+3)(n+2)(n+1)n(n-1)......6
_________________________________ S1
n (n-1)(n-2)......6*5*4*3*2
S1 = a1 =1
=> Sn = (n+4)(n+3)(n+2)(n+1)
____________________
5*4*3*2
an = Sn - Sn-1
(n+4)(n+3)(n+2)(n+1) (n+3)(n+2)(n+1)n
= ___________________ _ _________________
5*4*3*2 5*4*3*2
(n+3)(n+2)(n+1)
= _______________
30
聖誕節快樂XD
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