i need to prove 3|[(2^n-1)+1]?
912082
Your question is unclear.  What does your question equal to?
187387
(n)(3n-1)(3n-2).......(n+1)
483173
((6p - 18)/(9p)) / ((3p - 9)/(p^2 + 2p))\n((6(p - 3))/3(3p)) / ((3(p - 3))/(p(p + 2)))\n((2(p - 3))/(3p)) / ((3(p - 3))/(p(p + 2)))\n((2(p - 3))/(3p)) * ((p(p + 2))/(3(p - 3))\n(2p(p - 3)(p + 2))/((9p)(p - 3))\n(2p(p + 2))/(9p)\n(2(p + 2))/9\n(2p + 4)/9
920343
(4/3)[x^3/(2-x)]^(1/3)\n[(2-x)(3)x^2-x^3(-1)]/\n(2-x)^2=\n\n 4x[1/(2-x)]^(1/3)\n[3x^2(2-x)+x^3]/\n3(2-x)^2
919554
Factor out 3.\n3(x^2+2x+1)=0\n3(x+1)^2=0\nThe only answer then is x=-1 to make the expression equal 0.
846663
(x+3)(x-1)\n\n x= -3\nx=1
810638
b=a+1\nc=a+2\nd=a+3\ne=a+4\nf=a+5\ng=a+6\nh=a+7\na^2+d^2+f^2+g^2\n=a^2+(a+3)^2+(a+5)^2+(a+6)^2\n=4a^2+28a+70\nb^2+c^2+e^2+h^2\n=(a+1)^2+(a+2)^2+(a+4)^2+(a+7)^2\n=4a^2+28a+70\nhence proved
123892
3(x – 2)^2 – 3(x – 2) – 6\n3((x-2)^2 - (x-2) -2)\n3(x^2-4x+4-x+2-2)\n3(x^2-5x+4)\n\n3(x-4)(x-1)\n\nYay!!
751318
x^2-x-6>0\nor, x^2-3x+2x-6>0\nor, x(x-3)+2(x-3)>0\nor, (x-3)(x+2)>0\nor x>3 and x<2\n\nhope this helps :)
756805
x<2   (-infinity,2)\nx>3   (3, infinity)\nx<2 see first interval\nx>-2 (-2,infinity)
1182922
3^3 + 3 + 3/3 = 27+3+1 = 31
703982
1.\n3x + 6/((x + 1)(x + 2)) Factor 3 out of the top.\n[3(x + 2)]/[(x + 1)(x + 2)] x+2's cancel\n3/(x + 1) <-- This is simplified.\n\nI don't have the energy to do the other ones right now.  :/
771358
3x + 4 < x + 8\n2x < 4\nx < 2\n\n{x | x < 2}
944695
12 * (3-e^-3x)^3 *  e^-3x
817835
x^3-x^2+3x-1\nx+4&#92;x^4+3x^3-x^2+11x-4\n      -(x^4+4x^3)\n              -x^3-x^2\n            -(-x^3-4x^2)\n                       3x^2+11x\n                     -(3x^2+12x)\n                                   -x-4\n                                 -(-x-4)\n                                         0\nx^4+3x^3-x^2+11x-4=(x+4)(x^3-x^2+3x-1)
416859
There is a rule of exponents that says (x^a)(x^b) = x(a+b).  Use this in reverse:\n\n(3^2x) - 3^(2x - 1) = 18\n(3^2x) - [3^(2x)*3^(- 1)] = 18\n(3^2x) - (3^(2x))/3 = 18\n3(3^2x) - (3^2x) = 54\n2(3^2x) = 54\n(3^2x) = 27\nlog(3^2x) = log27\n2xlog3 = log27\nx = log27/2log3\n= 1.5
715296
(A)  Show base case is true:\n\n(2*1)^2 = (2*1)(2)(2*1 + 1) / 3\n4 = 4\n\nAssume the equality holds for n = k\nShow that it holds for n = k+1\n\n2^2 + 4^2 + 6^2 + ... + (2k)^2 + (2(k+1))^2\n= 2k (k+1) (2k + 1) / 3 + (2k+2)^2\n= 2k (k+1) (2k + 1) / 3 + (4k^2 + 8k + 4)\n= (2k^2 + 2k) (2k + 1) / 3 + (12k^2 + 24k + 12) / 3\n= (4k^3 + 6k^2 + 2k + 12k^2 + 24k + 12) / 3\n= (4k^3 + 18k^2 + 26 k + 12) / 3\n\n2(k+1)(k+2)(2k+3) / 3 \n= 2 (k^2 + 3k + 2)(2k + 3) / 3\n= 2 (2k^3 + 3k^2 + 6k^2 + 9k + 4k + 6) / 3\n= (4k^3 + 18k^2 + 26k + 12) / 3\n\n(B)  Show base case is true\n5 (2!) ≥ 3^2\n10 ≥ 9\n\nAssume it is true for all n < k, Show it is true for n = k+1\n\n5(k+1)! ≥  5 (k+1) k! ≥  5 * 3 * 3^k ≥ 5 * 3^(k+1)