t1=1
t2=1+4*3^0=5
t3=1+4*(3^0+3^1)=17
t4=1+4*(3^0+3^1+3^2)=53
t5=1+4*(3^0+3^1+3^2+3^3)=161
Next term is 1+4*(3^0+3^1+3^2+3^3+3^4)=485
2006-09-14 05:10:29
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answer #1
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answered by astrokid 4
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1, 5, 17, 53, 161, 324
1,5 ... 5 - 1 = 4 (4 * 3 = 12)
5, 17 ... 17 - 5 = 12 (12 * 3 = 36)
17, 53 ... 53 - 17 = 36 (36 * 3 = 108)
53, 161 ... 161 - 53 = 108 (108 * 3 = 324)
Hope this is right, was a challenge! Had to search the next for inspiartion and ways to solve for these answers.
Jennifer
2006-09-14 05:28:20
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answer #2
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answered by jennifermlayne 2
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i will see that the 1st, third and so on. numbers multiply by using 3 every time. i will see that the 2d, 4th and so on. numbers subtract 2 every time. which might supply a continuation to the sequence: 2 9 6 7 18 5 fifty 4 3 162 a million 486... with fifty 4 the for the asked seventh term. even however, there are an endless form of formulae for the guy words which might supply any value to the seventh term!!!
2016-12-12 08:19:45
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answer #3
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answered by Anonymous
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485. each next number is 3 times as much as is in between the previous numbers added to the last number. um...5-1=4. so there's 4 in between 1 and 5. then 17-5=12 so there's 12 in between 5 and 17 which is 3 times 4 and so on.
sorry, my math was off.
2006-09-14 05:16:36
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answer #4
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answered by practicalwizard 6
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485
2006-09-14 05:19:57
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answer #5
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answered by Anonymous
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485
2006-09-14 05:17:18
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answer #6
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answered by Charles B 4
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121
2006-09-14 05:11:05
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answer #7
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answered by Reggie 3
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485 smartass
2006-09-14 05:17:13
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answer #8
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answered by pollypureheart 4
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485
the difference between adjacent numbers in the series is 4*3^n where n is the n'th number in the series... just so that you know i'm not copying :-P
2006-09-14 05:16:16
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answer #9
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answered by Shane 2
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