Use induction to prove that for all integers n >= 1,  (2n)! < 2^n( n!)^2?
963823
LHS = (2n)!\nRHS = 2^n( n!)^2\n\nfor n=1,\n(2n)! = (2)! = 1.2 = 2\n2^n( n!)^2 = 2^1( 1!)^2 = 2\nSo LHS = RHS, condition is not satisfied.\n\nfor n = 2,\n(2n)! = (4)! = 1.2.3.4 = 24\n2^n( n!)^2 = 2^2( 2!)^2 = 16\nSo LHS > RHS\n\nYour problem has errors!!!\n\nI'll try to prove:\n (2n)! >= 2^n( n!)^2 .. .... (ii)\n\nfor n = 1, \n\n(2n)! = (2)! = 1.2 = 2\n2^n( n!)^2 = 2^1( 1!)^2 = 2\nSo LHS = RHS, condition is satisfied.\n\nlet, for any arbitrary integer m, (ii) is satisfied.\n\nSo, \n (2m)! >= 2^m(m!)^2 \n\nnow, if n = m+1, (ii) becomes,\n(2(m+1))! >=2^(m+1) ((m+1)!)^2  \n\nnow,\n(2(m+1))! = (2m+2))! = (2m+2)(2m+1) (2m)! \n= 2(m+1)(2m+1) (2m)! \n\n2^(m+1) ((m+1)!)^2  = 2.2^m  (m+1)^2 (m!)^2 \n\n\nwe have assumed, (2m)! >= 2^m(m!)^2 \nnow, multiplying by 2(m+1)(2m+1) we get,\n2(m+1)(2m+1) (2m)! >= 2^m 2(m+1)(2m+1) (m!)^2 \n\nnow, 2^m 2(m+1)(2m+1) (m!)^2  = 2.2^m (m+1)(2m+1) (m!)^2 \n\nagain 2m+1 > m+1\n\nso, 2(m+1)(2m+1) (2m)! >= 2^m 2(m+1)(m+1) (m!)^2 \n\nor,\n(2(m+1))! >=2^(m+1) ((m+1)!)^2  \n\n(ii) is true for n = m+1\n(Proved)
10571
50 is right.\n{i^2 -(-1)^i | i=1,2,...}
367692
If p(1) is true and, for each integer k>1 the statements  p(1),p(2),....p(k) imply the statement p(k+1) then p(n) is true for all.\n\nIn other words if a statement is true for the integers 1..k\nand you can prove it from those for k+1, than it is true for all integers.\n\nIt is a stronger version of normal mathematical induction where one use P(k) to prove P(k+1).
100604
anything like, n.0... , n.1..., n.2..., n.3..., n.4... will be $n.\n\nanything like, n.5... , n.6..., n.7..., n.8..., n.9... will be $n+1.
699840
let N =nth term\n\ntherefore, for n=1,2,3,.............\n\nN=6*1.2^(n-1)\n\ni hope that this helps
882612
2x^2 ln(x) - x^2\n\nVerify by differentiating\n2x + 4x ln(x) - 2x = 4x ln(x)
528695
x to the n + y to the n = z to the n where n is an integer? ... damn it's been so long I can't remember for sure. Some Brit solved the proof already.
999954
1!=1\n2!=2\n3!=6\n4!=24\n5!=120\n>5!=some number ending in zero, since everything would be a multiple of 5!, and 5! is a multiple of 10.\nThus your answer is 1+2+6+4 mod 10 or being simplified, 3.
859057
2+(n-1)2+1/2(n-1)(n-2)1\n= 2 + 2n - 2 + (1/2)(n^2 - 2n - n + 2)\n= 2n + (1/2)(n^2 - 3n + 2)\n= 2n + (1/2)n^2 - (1/2)(3n) + (1/2)2\n= 2n + (1/2)n^2 - (3/2)n + 1\n= (1/2)n^2 + (1/2)n + 1
1000878
First let I(k,n)=int (log x)^k x^n dx where the integral goes from 0 to 1. Do an integration by parts with u=(log x)^k x^(n-1) and dv=x dx\nto get that I(k,n)=-k/(n+1). By induction, we then obtain that\nI(n,n)=(-1)^n n!/(n+1)^n I(0,n)\n=(-1)^n n!/(n+1)^(n+1).\n\nNow, write x^(ax)=exp(ax log x)=\nsum 1/n! (ax log x)^n=\nsum (1/n!) a^n x^n (log x)^n.\n\nNow integrate. Since the sum converges uniformly, the interchange of the sum and integral gives\nint x^(ax) dx=sum (1/n!) a^n int x^n (log x)^n dx\n= sum (1/n!) a^n I(n,n)\n=sum (-1)^n a^n /(n+1)^(n+1)\nas desired.
50362
It usually does:\n\nlim n->inf 1^n = 1\n\nOn the other hand,\n\nlim n->inf (1 + 1/n)^n = e
922060
My suspicion upon seeing this question is that it can be solved using mathematical induction.  After all, couldn't we simply 'build' such a (maximally connected) graph starting with n=2 vertices and 'grow' it from there?\n\nRather than take on the proof myself -- mostly because I am unsure of the definition of 'components' -- I leave it as an exercise for you.  However, it bears noting that the formula shown is the commonly accepted "sum of the first n integers" formula used in almost every text on induction.\n\nGood luck!
670692
Good question! Easy to state but unless you are familiar with this type of problem (which I was not) it is tricky. \n\nYou really want to know? The answer is:\n\n573,147,844,013,817,084,100\n\nThis is how I started. It's easy to prove it's > 2^50\n\n100/2 = 50  So you can take 2 steps at a time anywhere from 0 to 50 times. 0 would be that you climbed the 100 steps 1 at a time. If we let r equal the number of times (from 1 to 50) that you take 2 steps at a time, and for the sake of simplicity we will assume that we can only take 2 steps either from the 0th step or from an even step, then for each value of r the number of distinct ways of doing it are given by the binomial coefficient nCr = n!/(n-r)!r!. So 50C1 = 50; 50C2= 1225; 50C3 = 19,600 etc..  \nNote: in reality there are 98 ways you can combine 1 two-step with 98 1-steps so 50C1 = 50 is too low. However,let's continue. \nWe have to sum up all of these coefficient values.\nBut this is not hard to do because the sum of nCr for r= 1 to n is 2^n (2 raised to the power of n). I am not going to prove this; that's already been done. but you get the idea from the following expansions:\n\n2^3 = (1+1)^3 = 1+3+3+1 =8\n2^4 = (1+1)^4 = 1+4+6+4+1 =16\n2^5 = (1+1)^5 = 1+5+10+10+5+1 =32 and so on...\n\nWhen we use (1+1)^n as a specific case of (a+b)^n we just get the binomial coefficients and we that they sum to a power of 2. So we know the answer is greater than 2^50.\nYou will be closer if you add 1+98C1+96C2+94C3+...+54C48+52C48+50C49+1\n\nBut there is still a problem..because you have to subtract some because of starting on an odd step..\n\nAnyways, I realized it was problem in restricted partition compositions so I went to this site and they had a calculator and I used it. That's how I got the answer lol!\n\nhttp://www.btinternet.com/~se16/js/partitions.htm
466636
Let me see : \n6! = 1 x 2 x 3 x 4 x 5 x 6\n5! = 1 x 2 x 3 x 4 x 5 \n6!/5! = ( 1 x 2 x 3 x 4 x 5 x 6 )/( 1 x 2 x 3 x 4 x 5 )=\n      = 1/1 x 2/2 x 3/3 x 4/4 x 5/5 x 6 =\n      = 1   x  1  x  1  x  1  x  1  x 6 \n6!/5! = 6
936579
your question seems to be incomplete.\nso far....\nwidth=x\nlength= 2x-1
432827
(n-1)!/(n+1)! = (n-1)!/((n+1)n(n-1)!)\n= 1/((n+1)n)\n= 1/(n^2+n)
360700
a = 1, b = 2, n = 3:\n\n1^2 = 1 = 4 = 2^2 (mod 3) but clearly 1 != 2 (mod 3)