數學--數列的問題

2009-09-28 12:58 am
1.已知一個數列9,99,999,9999......
(a)求該數列的通項T(n)
(b)求T(1)+T(2)+ T(3)+....+T(n),答案以n表示

2.對於級數(1)+(2+2^2)+(2^3+2^4+2^5)+....,求
(a)第n個括號內的首項和末項
(b)第n個括號內所有項之和
(c)首n個括號內所有項之和

3.已知S1=a+ar+ar^2+...+ar^(n-1)及S2=a^2+(a^2)(r^2)+(a^2)(r^4)+.......+(a^2)(r^(2n-1)),證明(r+1)S2=(r-1)S1^2+(2a)(S1)
更新1:

第2-3條點計?

回答 (1)

2009-09-28 7:21 am
✔ 最佳答案
(1a) T(n) = 10^n - 1
(1b) T(1) + T(2) + ... + T(n)
= (10 - 1) + (10^2 - 1) + ... + (10^n - 1)
= (10 + 10^2 + ... + 10^n) - n
= 10(10^n - 1)/(10 - 1) - n
= 10(10^n - 1)/9 - n
(2a)第一個括號內有一項
第二個括號內有二項
第n-1個括號內有n - 1項
從第一個到第n-1個括號共有1 + 2 + 3 + ... + (n - 1)項
= n(n - 1)/2項
第n個括號內的首項 = 2^[n(n - 1)/2]
第n個括號內共n項,
末項 = 2^[n(n - 1)/2 + n - 1]
= 2^[(n - 1)(n + 2)/2]
(2b)第n個括號內所有項之和
= 2^[n(n - 1)/2] + ... + 2^[(n - 1)(n + 2)/2]
= 2^[n(n - 1)/2][1 + 2 + 2^2 + ... + 2^(n-1)]
= 2^[n(n - 1)/2][(2^n - 1) / (2 - 1)]
= 2^[n(n - 1)/2](2^n - 1)
(2c)首n個括號內所有項之和
= 1 + 2 + 2^2 + ... + 2^[(n - 1)(n + 2)/2] 共 k = (n - 1)(n + 2)/2 + 1項
= 1(2^k - 1)/(2 - 1)
= 2^k - 1
k = (n - 1)(n + 2)/2 + 1
= [(n^2 + n - 2) + 2]/2
= n(n + 1)/2
首n個括號內所有項之和 = 2^[n(n + 1)/2] - 1
3. 先簡化S1, S2
S1 = a + ar + ar + ... + ar^(n - 1)
= a(r^n - 1)/(r - 1)
S2 = a^2 + a^2r^2 + a^2r^4 + ... + a^2r^[2(n - 1)]
= a^2(r^2n - 1)/(r^2 - 1)
= a^2[(r^n)^2 - 1]/[(r - 1)(r + 1)]
= a^2[(r^n - 1)(r^n + 1)]/[(r - 1)(r + 1)]
左邊 (r + 1)S2 = a^2(r^n - 1)(r^n + 1) / (r - 1)
右邊 (r - 1)*S1^2 + 2aS1
= (r - 1)a^2(r^n - 1)^2/[(r - 1)^2] + (2a)a(r^n - 1)/(r - 1)
= a^2(r^n - 1)^2/(r - 1) + 2a^2(r^n - 1)/(r - 1)
= [a^2(r^n - 1)/(r - 1)][(r^n - 1) + 2]
= a^2(r^n - 1)(r^n + 1) / (r - 1)
= 左邊


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