The general term
What is the general term of
7,13,25,49,97
2007-02-17 13:29:22 · 26 個解答 · 發問者 ? 1 in 科學 ➔ 數學
general term
3*2^n+1
1st term=3*2+1=7
2nd term=3*4+1=13
3rd term =3*8+1=25
4th term=3*16+1=49
5th term=3*32+1=97
2007-02-18 07:10:38 · answer #1 · answered by Tiana 2 · 0⤊ 0⤋
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2007-03-20 19:41:06 · answer #2 · answered by ? 5 · 0⤊ 0⤋
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2007-03-20 14:17:11 · answer #3 · answered by ? 5 · 0⤊ 0⤋
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2007-03-20 13:59:36 · answer #4 · answered by ? 4 · 0⤊ 0⤋
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2007-03-20 13:11:48 · answer #5 · answered by - 【分開簡單】MaTtHeW L【抹去往事極難】 5 · 0⤊ 0⤋
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2007-03-20 01:10:18 · answer #6 · answered by 大地小羊 6 · 0⤊ 0⤋
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2007-03-19 23:08:15 · answer #7 · answered by ? 5 · 0⤊ 0⤋
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2007-03-19 21:42:21 · answer #8 · answered by 知識21號 6 · 0⤊ 0⤋
T(1) = 7
T(2) = 13 = T(1) + 6 = 7 + 6
T(3) = 25 = T(2) + 12 = 7 + 6 + 2(6) = 7 + (1 + 2)(6) = 7 + (2^0 + 2^1)(6)
T(4) = 49 = T(3) + 24 = 7 + (1 + 2)(6) + 4(6) = 7 + (1 + 2 + 4)(6) = 7 + (2^0 + 2^1 + 2^2)(6)
T(5) = 97 = T(4) + 48 = 7 + (1 + 2 + 4)(6) + 8(6) = 7 + (1 + 2 + 4 + 8)(6) = 7 + (2^0 + 2^1 + 2^2 + 2^3)(6)
....
T(n) = 7 + [2^0 + 2^1 + 2^2 + ... + 2^(n-2)](6)
Consider x = 2^0 + 2^1 + 2^2 + ... + 2^(n-2),
x = 2^0 + 2^1 + 2^2 + ... + 2^(n-2) ..... (1)
2x = 2^1 + 2^2 + ... + 2^(n-2) + 2^(n-1) ..... (2)
(2) - (1),
x = 2^(n-1) - 2^0 = 2^(n-1) - 1
Therefore, T(n) = 7 + [2^(n-1) - 1](6) = 7 + 6[2^(n-1) - 1].
2007-02-18 21:06:14 補充:
不好意思, 太大意了, 還沒把答案約簡, Dave Kwok ~ Totti的答案是對的.T(n)= 7 6[2^(n-1) - 1]= 7 6 * 2^(n-1) - 6= 3 * 2 * 2^(n-1) 1= 3 * 2^n 1
2007-02-18 21:07:15 補充:
不好意思, 太大意了, 還沒把答案約簡, Dave Kwok ~ Totti的答案是對的.T(n)= 7 十 6[2^(n-1) - 1]= 7 十 6 * 2^(n-1) - 6= 3 * 2 * 2^(n-1) 十 1= 3 * 2^n 十 1
2007-02-17 13:51:22 · answer #9 · answered by 琴生 5 · 0⤊ 0⤋
T(2)-T(1)=13-7=6
T(3)-T(2)=25-13=12
T(4)-T(3)=24
Which implies T(n+1)-T(n)=6(2)^(n-1)
T(2)=T(1)+6
T(3)=T(1)+6+6(2)
......
T(n)=T(1)+6+6(2)+......+6(2)^(n-1)
=T(1)+6[2^(n-1)-1]
=7+6[2^(n-1)-1]
2007-02-19 13:19:52 補充:
實際上7+6[2^(n-1)-1]也不是不對......
2007-02-17 13:39:53 · answer #10 · answered by 打倒美帝紙老虎 6 · 0⤊ 0⤋