given the geometric sequence:\n0,1,8,27,64,125.\nfind the pattern for the nth term.\n\nis it n=1^3?
215925
0 = 0^3\n1 = 1^3\n8 = 2^3\nand so on..\n125 = 5^3\n216 = 6^3\n\nSo it's 216
518625
okay so what N means is that you are plugging in any integer value for n. integers, as a reminder, a whole, countable numbers like 1, 6, 88, 45798.\nSo teh first four integers (values of n) are obvious 1, 2, 3, 4\nso the sequence is\n(2*1 -5), (2 *2 -5), (2 *3-5), (2 *4-5)\n-3, -1, 1, 3.\n\nHere's one way to prove that it's arithmetic:\nYou can graph them as continuous variables along an x y coordinate system. That is, graph the points (1, -3), (2, -1), (3, 1), and (4, 3). If you connect them, you'll get a straight line. It's no surprise that the equation of that line (if you choose to find it) is y=2x-5. Because the points lie on a straight line, it's arithmetic (as opposed to geometric).
475819
We are asked to write an equation, and no answers so far have any equation.\n\nIn each of the equations, we will write a function that will give us the nth term in the sequence. The general procedure is to note the relationship between the successive terms in the series. In the first one, the series goes up by 2 when n goes up by 1. So the formula will need a term 2n in it somehwere. Then we have to consider the initial value of the series, so that for n = 1 we get the correct result.\n\n1. T = 2n - 1\n\nThe second series is going to need a 3n term somewhere, but the series is -5 for n = 1.\n\n2. T = 3n - 8\n\nThe fifth series is a little interesting, in that we add x and subtract 1 for each successive term. That means we will need n(x - 1) in the equation, but we have to make the initial value work.\n\n3. n(x - 1) + 3\n\nMatt Z, are you THE Matt Z?
219492
Multiply the last 2 terms together and then add 1.\nIf that is the pattern (looks like it is) then it should be 155, not 153.\nTo find the next 3 terms, just keep going. (You'll need a calculator with lots of digits).\n22 * 155 + 1 = 3411\n155 * 3411 + 1 = 528706\n3411 * 528706 + 1 = 1803416167
269915
Arithmetic sequence:  1, 2, 3, 4, 5, 6, 7\nGeometric sequence:  1, 2, 4, 8, 16, 32, 64
612255
1113213211\n\nAfter starting the sequence with 1, each term in the sequence consists of groups of two numbers based on the previous term - the first being the quantity and the second specifying which digit.
19211
There is no closed form for the nth derivative of this function.  I symbolically calculated and factored the first 50 derivatives on MathCAD, and there was not a pattern.  Sorry, but you're out of luck....
495572
Bionomial expansion\n\nthis has got n+1 terms\n\nthe terms are (nCk) a^kb^n-k where k varies from 0 to n\n\nfor example (a+b)^3\none term (3C0) a^0b^3 = b^3\nanother term (3C1)a^1b^2 = 3ab^2\nanother (3C2)a^2b^1 = 3a^2b\nanother (3c3)a^3b^0 = a^3\n\nadd them and get a^3+3a^2b+3ab^2+ b^3\n\n(nCk) = n!/(k!(n-k)!)\n\nThe n'th line of the pascal triangle gives you the coeffiecients . Starting from 0.
477152
... 27, 26, 676, 678, 677, 458329, 458331, 458330, ...\n\nThis is sequence A059417 in the Online Encyclopedia of Integer Sequences.\n\nSquare, Add 2, Subtract 1, repeat...
369457
The nth term is n cubed.\nThe third tem is 27.
508515
That last number should be 21.\n\nThat sequence is called the Fibonacci sequence.  It's the most famous sequence probably ever.\n\nThe 22nd is 17711.  You can look it up in table...or there's actually a way to compute it without having to do all the ones up to 22.\n\nThe formula for the nth Fibonacci number is:\n\nF(n) = 1/sqrt(5) [((1+sqrt(5))/2)^n + ((1-sqrt(5))/2)^n]\n\nKind of messy, but easier than writing them all out up to 22.
465717
Hello dear Asker ;\n\n{ as you know ;\nP(n , r) = n! / ( n –  r )! }...\n\nso ;\n\nP (5 , 3) = 5! / ( 5 – 3)! \n=5! / 2! \n= (5 * 4 * 3 * 2 * 1) / (2 * 1) \n= 120 / 2 \n= 60 \nTotally We cane make 60 words with out any repetition\n\nGood Luck dear
397207
21,\n52,\n40,\n80,\n104\n\neach pattern, you are adding a number in the one before the number\n1 + 0 = 1, 1 + 1= 2, 2 + 1 = 3, 3 + 2 =5, 5 + 3 = 8, and so on and so forth\n\nsorry if my answer is wrong im just a kid trying to answer correctly your question....\n\nand im very sorry if my words are illiterate!!
518610
The nth term is the 'method of differences'. If you have a sequence then you can usually use the nth term to figure out the formula for that sequence. There are linear (normal) and quadratic sequences (to do with something squared) and I'll give an example of the first, if you'd like the second explained, I will do that too (just edit you question, then I'll answer it)-I just don't want to confuse you with too much information. \n\nEg. 2, 5, 8, 11...\n-You can see that this goes up in 3s so the first part of your formula is 3n. \n-So you put 3n and work out that sequence. (n stands for the term in the sequence, eg. 2-n=1, as 2 is the first term in the sequence). So n would be 1, 2, 3, 4... But it's 3n so you must times those by 3, so now: 3, 6, 9, 12...\n-Now we can put the two next to each other and work out the difference to get the final part of our formula. \n2, 5, 8, 11...\n3, 6, 9, 12... \n\nWe can see that we have to minus one from the bottom sequence to get to the top sequence so our formula for this sequence is: 3n-1.\n\nHope this has helped!!
356308
180, 294, ...\n\nIt is the place of the number squared times the place of the previous number; (n^2)(n-1)\n\nYou can get the last one I hope!
245841
Hi! This recurrence relation can be evaluated for n=5^s, s>0, since we have been given T(1).\n\nUsing the given formula,\n\nT(625)=7*T(125) + 10*625\n= 7*[ 7*T(25) + 10*125 ] + 10*625\n= 7*[ 7*{ 7*T(5) + 10*25 } + 10*125 ] +10*625\n= 7*[ 7*{ 7*( 7*T(1) + 10*5 ) + 10*25 } + 10*125 ] +10*625\n\nNow, T(1)=1\n\nInstead of evaluating the expression, open the brackets, to get,\n\nT(625) = 7^4 + 7^3*10*5 + 7^2*10*25 + 7^1*10*125 + 10*625\n\nwhich can also be written as:\n\nT(625) = (7^4) + (7^3)*10*(5^1) + (7^2)*10*(5^2) + (7^1)*10*(5^3) + (7^0)*10*(5^4)\n\nSo, in general for n=5^s, we can also express the relation as,\n\nT(5^s) = 7^s + Σ{7^(s-k-1)*10*5^(k+1)} where k varies from 0 to (s-1), provided s>0.\n\nThe value of T(625) from the expression above comes out to be: 46801
541278
Take each number, multiply by two, and then subract two.\n\n3*2 - 2 = 4\n4*2 - 2 = 6\n6*2 - 2 = 10\n10*2 - 2 = 18\n\nedit: I'm sure that the two possible answers are related.  I would also get 34, then 66, then 130 as the next numbers in the sequence.  It has to do with the fact that it started with three.  If we had started at 2, then my system would always yield 2 every time, but you guys wouldn't.  I think both are acceptable, although I will admit that the other method is more interesting.