The sequence goes 3,7,25,121.... Thats all I have to work with... I need to find the pattern and the 10th term. Thanks ahead of time.
2007-12-14 04:23:44 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
the sequence should be
T(n) = n! +1 .................. ! means factorial
but the first term is 3 = 2!+1
so you probably is having
T(n) = (n+1)! +1
the tenth term would be
11! + 1 = 39 916 801
2007-12-14 04:37:36 · answer #1 · answered by KDFC 3 · 1⤊ 0⤋
The "Online Encyclopedia of Integer Sequences" can be really helpful for questions like these.
http://www.research.att.com/~njas/sequences/A038507
http://www.research.att.com/~njas/sequences/A000142
Notice, if you subtract 1 from each number you get:
2, 6, 24, 120...
If this had started with 1, you would have:
1, 2, 6, 24, 120...
Each number is n times the previous number (or another way to say it is n!)
1, 2, 6, 24, 120, 720, 5040, 40320, 362880, 3628800, 39916800
But you don't have the first term:
2, 6, 24, 120, 720, 5040, 40320, 362880, 3628800, 39916800
And your sequence is one more than each of these numbers:
3, 7, 25, 121, 721, 5041, 40321, 362881, 3628801, 39916801
A(n) = (n+1)! + 1
A(10) = 11! + 1
A(10) = 39,916,801
2007-12-14 04:53:04 · answer #2 · answered by Puzzling 7 · 0⤊ 0⤋
n!+1 is the only answer
2007-12-14 05:02:02 · answer #3 · answered by Ahmed Zia 3 · 0⤊ 0⤋
Danny is correct
2007-12-14 04:50:37 · answer #4 · answered by Pranil 7 · 0⤊ 0⤋
2,3,7,25,121,?
2015-12-06 06:20:20 · answer #5 · answered by Roza 1 · 0⤊ 0⤋