master_q 0 Posted September 28, 2003 (edited) {Official} Star Trek Trivia #229 Score One For the Team! Oh, NO! A math question! 12*; 36; 108; 324; . . . . . . What is the next #? [i want to know what the next five #s are. I should have a total of 5 replies giving a total of 5 answers.] BONUS POINTS: If someone can give me a standard equation for this kind of sequence, then the team will get an additional 3 points! {EDIT} *LOL. I think I need to take a typing lesson. The numbers listed now should be correct Master Q StarTrek_Master_Q@yahoo.com Edited September 28, 2003 by master_q Share this post Link to post Share on other sites
Theunicornhunter 2 Posted September 28, 2003 12*; 36; 108; 324; . . . . . . What is the next #? 972 Share this post Link to post Share on other sites
Indy 8 Posted September 28, 2003 Next number - 2,916 Share this post Link to post Share on other sites
removed 0 Posted September 28, 2003 (edited) 7th Sequence = 8 748 This equation works, but you have to start the variable x at 0 for it to work: (click on spoiler) Click for Spoiler: equation: n = 12(3^x) ex1. n = 12(3^0) n = 12 ex2. n = 12(3^1) n = 36 ex3. n = 12(3^2) n = 108 ex4. n = 12(3^3) n = 324 Edited September 28, 2003 by bysty Share this post Link to post Share on other sites
master_q 0 Posted September 28, 2003 7th Sequence = 8 748 This equation works, but you have to start the variable x at 0 for it to work: (click on spoiler) Click for Spoiler: equation: n = 12(3^x) ex1. n = 12(3^0) n = 12 ex2. n = 12(3^1) n = 36 ex3. n = 12(3^2) n = 108 ex4. n = 12(3^3) n = 324 equation: n = 12(3^x) I don't like that equation. I want the first term to be 1 and not 0. (x should = 1 and give us an answer of 12. When I plug x=2 I should get 36 and so on) So when n = 12 (our first term) I want to be able to plug 1 into the equation to get 12 You are close. Just think about it. Master Q StarTrek_Master_Q@yahoo.com Share this post Link to post Share on other sites
Captain Bolivar 0 Posted September 28, 2003 The next number is 26,244 Share this post Link to post Share on other sites
Captain Bolivar 0 Posted September 28, 2003 Here is a general equation that gives 12 for an x of 1. n = 12[3^x - {(2)(3^x-1)}] note: (3^x-1) is NOT (3^x)-1 (3^x-1) IS 3 to the exponent (x-1) SUB IN 1 FOR x 12 = 12[3^1-{(2)(3^1-1)}] 12 = 12[3 - {(2)(3^0)}] 12 = 12[3 - {(2)(1)}] 12 = 12[3 - 2] 12 = 12[1] 12 = 12 SUB IN 3 FOR x 108 = 12[3^3 - {(2)(3^3-1)}] 108 = 12[27 - {(2)(3^2)}] 108 = 12[27 - {(2)(9)}] 108 = 12[27 - 18] 108 = 12[9] 108 = 108 Share this post Link to post Share on other sites
ensign_beedrill 0 Posted September 29, 2003 Fifth number - 78,732 Share this post Link to post Share on other sites
master_q 0 Posted September 29, 2003 (edited) Here is a general equation that gives 12 for an x of 1.n = 12[3^x - {(2)(3^x-1)}] note: (3^x-1) is NOT (3^x)-1 (3^x-1) IS 3 to the exponent (x-1) SUB IN 1 FOR x 12 = 12[3^1-{(2)(3^1-1)}] 12 = 12[3 - {(2)(3^0)}] 12 = 12[3 - {(2)(1)}] 12 = 12[3 - 2] 12 = 12[1] 12 = 12 SUB IN 3 FOR x 108 = 12[3^3 - {(2)(3^3-1)}] 108 = 12[27 - {(2)(3^2)}] 108 = 12[27 - {(2)(9)}] 108 = 12[27 - 18] 108 = 12[9] 108 = 108 The equation seems to work, but I don't like it being that long. In certain cases you can't avoid a long equation, but if you can then try to make it as small as possible. It works and I guess I'll give credit, but I can think of a much smaller equation that would do the same job. Look Bysty's equation. It is not right, but it is close. n = 12(3^x) n = 12(3^1) = 36 n = 12(3^2) = 108 . . . Bysty said; This equation works, but you have to start the variable x at 0 for it to work Now think about that! If you have to "start the variable x at 0," then how do you think you can bump that so you can start with 1? Master Q StarTrek_Master_Q@yahoo.com Edited September 29, 2003 by master_q Share this post Link to post Share on other sites
removed 0 Posted September 29, 2003 (edited) I think I have it. equation: n = 12(3^(x-1)) ex1. n = 12(3^(1-1)) n = 12 ex2. n = 12(3^(2-1)) n = 36 ex3. n = 12(3^(3-1) n = 108 ex4. n = 12(3^(4-1) n = 324 Edited September 29, 2003 by bysty Share this post Link to post Share on other sites
Captain Bolivar 0 Posted September 30, 2003 That is much nicer! It's sweet and to the point. Well, it's been a few years since I've taken math; i'm not accustomed to thinking in mathematical terms. I'm actually impressed with myself for creating my huge and almost completely redundant equation. Anywho, Well done Bysty! Share this post Link to post Share on other sites
removed 0 Posted October 2, 2003 Thanks Commander Bolivar. I used your equation to help me out. I hadn't even thought of the whole subtracting an exponent thing. I was thinking a bit too complex. Lol. Share this post Link to post Share on other sites
master_q 0 Posted October 2, 2003 (edited) Good Job Everyone! +17 Points Master Q StarTrek_Master_Q@yahoo.com Edited October 2, 2003 by master_q Share this post Link to post Share on other sites