Binomial Theorem...Please HELP!!?
971178
You can always myltiply all the terms that means ex.\n(2-x)^3=(2-x)*(2-x)*(2-x) \nBut that quite a task if the expression is powered to more than 2-3\n\nThen you have the Binomial Formula for Positive integral n\n. . . . . . k=n\n(x+y)^n=∑ C(n,k)*(x^(n-k)*y^(k) ;  (%%%%)\n. . . . . . k=0\n\nAnd C(n,k)=n!/[k!*(n-k)!]. .. . . . . . (######)\n\nC(n,k) is called the binominial Coefficient\n0!≡ 1 (Definition)\nn!= n* (n-1)*......1\n\nnow your problem is  reduced to find the binomial Cofficients for the expression (%%%%)\n\nI will show you one abitray example:\nC(6,4)=6!/[4!*(6-4)!]=\n6 * 5 * 4 * 3 * 2 * 1 /[(4*3*2*1)*(2*1)]=\n720/(24*2)=15\n\nyou can also use \nPascal triangel to find the Binomial Coefficients\n\nn k →\n↓\n: . . . . . . . . . . . . . . .1\n. . . . . . . . . . . . . .  1. 1\n. . .. . .. . . . . . . . .1. 2 . 1\n. . . . . . . . . . . . .1. 3 . 3 . 1\n. . . . . . . . . . . .1. .4. 6. .4. 1\n. . . . . . . . . . .1. .5.10.10. 5 .1\n. . . . . . . . . .1.. 6.15.20 15 6. 1\n\nIt's easy to construct the triangel of Pascal, \nyou see the number under two other numbers is the sum of these numbers. It's is somtime helfuld to construct a P- triangel, if you don't have a calculator or a formula handbook, because you can also look the coefficient up in a table.\n\nTake the example (1/3x - x)^6.we find by use of pascal T: the binomial coefficient to bee\n1.. 6.15.20 15 6. 1 (the last line), \nand we have from the formula (%%%%)\n\n. . . . . . k=6\n(1/3x - x)^6=∑ C(6,k)*(x^(6-k)*y^(k) ;  (%%%%)\n. . . . . . k=0\nso C(6,0)=1, C(6,1)=6, C(6,2)=15, C(6,3)=20 etc\n\nand your solution to this problem would be see (%%%%)\n1 * (1/3x)^6 + 6*(1/3x)^5(-x)^1 + 15(1/3x)^4(-x)^2 + 20(1/x)^3(-x)^3 +15(1/3)^2(-x)^4 + 6^(1/3)^1(-x)^5 + 1*(-x)^6\n\nConclusion\n1) you can alway do the hard task and calculate the oldfashion way\n2) you can use Binomil formula and get the binomial coefficients 3 different way:\n. . . . . . A) use Pascal Triangel\n: . . . . . B) calculate by use of (#####)\n: . . . .  .C) or simply look the coefficient up in a formula\n. . . . .  . . . .book.
703053
so the Q is basically asking for two binomials which, if multiplied will end up as a polinomial...\nAs a matter of fact any two binomials can form a polynomial.\n(2X+3) (x-7)...this is an e.g.
1030587
I like to use C(n,k) for the combiniation of n things taken k at a time.\nUse the binomial theorem to expand (1+k)^n\n\n(1+k)^n = C(n,0) (1^n)k^0 + C(n,1)(1^(n-1))k^1 + C(n,2)(1^(n-2))k^2 + ... + C(n,n-1)(1^1)k^(n-1) + C(n,n)(1^0)k^n\n\n1 ^ of anything is 1 , and if we set k=1, then we get\n\n(1+1)^n = 2^n = C(n,0) + C(n,1) + C(n,2) + ... + C(n,n-1) + C(n,n)\n\nif this sum =2048, then 2^n=2048 or n=11
943885
if you take the numbers one-by-one: 25! / ( (25-1)!*1! )\n\ncombination from n numbers taken by k:  n! / ( (n-k)!*k! )\n\nfor example:\n    -25 numbers, taken by 7: 25! / (18!*7!)\n    -25 numbers, taken by 4: 25!/ (21!*4!)
204914
I only need four 0.\nbinary:0!+0!+0!+0!=100
885905
Yes, and this is because of Binomial theorem.\n\nAccording to Binomial theorem\n(a+b)^n = sum(k=0 to k=n)[(a^k) * (b^(n-k))]\n\nSo, with each term in your expantion, we observe 3 things\n\n1. Power of one of the variable increases from zero to n.\n2. Power of remaining of the variable decreases from n to zero.\n3. The coefficient of each term from starting to last will have coefficient (n perm k), where n is the power of (a+b) on RHS and k is the number of the term you are evaluating.\n\nI hope this answers your question.\n\nAll the best.
823097
Can't prove it as n cannot = + or -1 and k be an interger.\nIt appears to work with other odd integers. Change your question then it may be possible to prove it.
862170
I'm sorry but your answer is wrong...let's review some basics. \n\nYou should know that, when multiplying 2 numbers: \n\npositive x positive = positive answer\nnegative x negative = negative answer\npositive x negative =  negative answer\n\nAlso, the FOIL rule is: \n\nFirst ("first" terms of each binomial are multiplied together) \nOuter ("outer" terms multiplied) \nInner ("inner" terms multiplied) \nLast ("last" terms multiplied) \n\nThe answer is then the sum of the terms obtained. Thus the general form is\n\n(a + b)(c + d) = ac (first) + ad (outer) + bc (inner)+ bd (last)\n\nWith this in mind, you can work out the problem: \n(4x + 5)(x + 1) = 4x(x) + 4x(1) + 5(x) + 5(1) = \n\n4x^2 + 9x + 5 is the Answer \n\nGood luck and practice! ;-)
1139003
(n/(n - 3)) + (5/(2n - 6))\n(n/(n - 3)) + (5/(2(n - 3)))\nMultiply everything by 2(n - 3)\n(2n + 5)/(2n - 6)
487838
1) Assuming an aritmetic sequence\n\nbn = b1 + (n-1)d \nletting b3=a1 = 20, then b7=a5 = 32, and b25 = b23\na5 = a1 + (5-1)d\n32 = 20 + (5-1)d\n32 = 20 + 4d\n12 = 4d\n3 = d\n\nb25 = a23 = 20 + (23-1)*3\nb25 = 20 + 22*3\nb25 = 20 + 66\nb25 = 86\n=====================================================\n2)\na75 = a1 + (75-1)d; d=-3\n\na75 = 5 + (75-1)*-3\na75 = 5 + 74*-3\na75 = 5 + -222\na75 = -217\n\nSn = n/2(a1+an)\nS75 = 75/2(5 + -217)\nS75 = 75/2(-212)\nS75 = -7950\n===================================================\n3) ????\n\na5 = 5*5 - 4 = 21\na15 = 5*15 - 4 = 71\na25 = 5*25 - 4 = 121\n\nOR MAYBE\n\na1 = 5*1 - 4\na1 = 5 - 4\na1 = 1\na30 = 5*30 - 4\na30 = 150 - 4\na30 = 146\n\nSn = n/2(a1+an)\nS30 = 30/2(1 + 146)\nS30 = 30/2(147)\nS30 = 2205\n================================================\n4)\n\nrth term is nCr-1m^n-(r-1)k^r-1\n\n5th term =\n4C4m^4-(5-1)k^5-1\n4C4 = 1\n\n5th term = k^4
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.
929249
How do you know the answer is 96?  Wouldn't you need to use the P(n,r), or 6 choose 3?  \n\nThe formula (if that is correct) would be n!/(n-r)!  \n\n6x5x4x3x2 divided by (6-3)!.  This would equal 120.\n\nBut I don't know.  I'm taking a damn discrete math course myself and we're working on the same junk!  good luck!
557990
This is a binomial problem.  The number of trials is 4 and the probability of success is 1/6.  No more than twice means either 0, 1, or 2 sixes.  So\n\n4!/(0!4!)(1/6)^0(5/6)^4 + 4!/(1!3!)(1/6)^1(5/6)^3 + 4!/(2!2!)(1/6)^2(5/6)^2\n\n= 0.482 + 0.386 + 0.116\n\n= 0.984
548405
please dont hate math\nshow interest\nthen you do well\n\nFOIL\nFirst-Out-Inner-Last\n\nThis method is used for multiplying two binomials.\nsay (a+b)(x+y)\n\n( a + b ) (x + y)\n\nFirst------first terms in the two binomials. they are to be multiplied. so ax\nthen Outer a and y. so ay\nInner-----bx\nLast-----last terms in the binomials. so 'by'\nthen add all the terms.\nax+ay+bx+by\n\nyou may consult me
812449
1/n! - 1/(n+1)! =  (n+1)!/[(n!)(n+1)!] - n!/[(n!)(n+1)!]\n                        = 1/[(n!)(n+1)!] X {(n!)(n+1-1)}\n                        = n*n!/[(n!)(n+1)!]                                 (divide by n)\n                        = n!/[(n-1)!(n+1)!]                  (divide by n!)\n                        =  1/[(n-1)!(n+1)]
205892
n! =n*n-1*n-2...until you get to zero.\n\nso 4! equals 4*3*2*1\nwhich is 24
192872
You must look at the formula for standard deviation of the sample distribution of (I'm thinking) binomial distributions. The proportion "p" is a parameter of the formula. You should check it out but, if I remember correctly, it's:\ns.d. of the sample distribution = sqrt[p*(p-1)/n]