請教一題數學 有關遞迴數列

(1) S_1 = (4/3)a_1 - (1/3)2^(n + 1) + 2/3

Sub. n = 1, a_1 = (4/3)a_1 - 4/3 + 2/3 => a_1 = 2

(2) S_n - S_n - 1

= (4/3)(a_n - a_n-1) - (1/3)(2^n)

a_n = 4(a_n-1) - 2^n

x_h = 4^n, For x_p, let a_n = x*2^n

x*2^n = 4x(2^(n-1)) - 2^n

2^n(x - 2x - 1) = 0 => x = 1

General solution: a_n = 4^n - 2^n

(3) S_n

= (4/3)(4^n - 2^n) - (1/3)(2^(n+1)) + 2/3

= 4^(n + 1)/3 - 2^(n + 2)/3 - 2^(n + 1)/3 + 2/3

= (2/3)(2^(2n + 1) - 2^(n + 1) - 2^n + 1)

T_n

= (3/2)[1/(2^(n + 1) - 3 + 1/2^n)]

As Σ T_n

= (3/2) Σ [1/(2^(n + 1) - 3 + 1/2^n)]

< (3/2) Σ [1/2^(n + 1) ] (考慮無限級數)

< 3/2


2011-05-07 09:43:56 補充：
唔駛歹勢啦。其實只要將a_n的式子代入S_n就可以了。

圖片參考：http://imgcld.yimg.com/8/n/AC06918685/o/151105060881513872136371.jpg



圖片參考：http://imgcld.yimg.com/8/n/AC06918685/o/151105060881513872136382.jpg
a1=S1=(4/3)a1-4/3+2/3= > (1/3)a1=2/3= > a1=2an=Sn-S(n-1)=[(4/3)an-(2^(n+1))/3+2/3]-[(4/3)a(n-1)-(2^n)/3+2/3]=(4/3)an-(4/3)a(n-1)- (2^n)/3= >(1/3)an =(4/3)a(n-1)+ (2^n)/3= > an=4*a(n-1)+ 2^n=2^2* a(n-1)+ 2^n …($)We claim that an=2^(2n)-2^n…($$)prove by induction.a1=2^2-2^1=2, done,if an=2^(2n)-2^nthenfrom ($)a(n+1)=2^2*an+2^(n+1) { from ($$) }=2^2*(2^(2n)-2^n)+2^(n+1)=2^(2n+2)-2^(n+1)=2^(2(n+1))-2^(n+1), done.Sn=(4/3)*an-(2^(n+1))/3+2/3=4/3*(2^(2n)-2^n)- (2^(n+1))/3+2/3=2/3*[2^(2n+1)-3*2^n+1]=2/3*[(2^(n+1)-1)*(2^n-1)]Tn=2^n/Sn=3/2*[2^n/((2^(n+1)-1)*(2^n-1))=3/2*[1/2^n-1)-1/(2^(n+1)-1)]ΣTn=Σ3/2*[1/2^n-1)-1/(2^(n+1)-1)]{this is a telescope series and lim 1/(2^(n+1)-1)=0, hence : }=3/2*(the first term 1/(2^1-1))=3/2*(1)=3/2[[Done]] To ： myisland8132( 知識長 )= (3/2) Σ [1/(2^(n + 1) - 3 +1/2^n)] < (3/2) Σ [1/2^(n + 1) ]The inequality is wrong, thecorrect form is(3/2) Σ [1/(2^(n + 1) - 3 + 1/2^n)]> (3/2) Σ [1/2^(n+ 1) ]{Considerthe sigma part, for n=1, left=1/(1+1/2)=2/3,and right=1/4,left > right.In general,1/{(2^(n +1) - 3 + 1/2^n) }=1/[2^(n +1)-(positive)].> 1/2^(n+ 1). }So I think your proof is also wrong.

2011-05-07 14:58:02 補充：
老王 ( 大師 5 級 ) 的才對，
我的回答和老王一樣，好像多餘了。哈!
不過，寫很久，捨不得刪。想留一個紀念。歹勢!

好像是這題
http://tw.knowledge.yahoo.com/question/question?qid=1010040104308

歹勢
只有解出前兩題，
http://tw.myblog.yahoo.com/sincos-heart/article?mid=2235&prev=1185&next=2230
第3題不知要用什麼技倆。

2011-05-07 17:09:22 補充：
第三題老王大大的解法應該比較正確