Mathematical Induciton help needed?
926069
suppose it is true for some n . I do not know because you say 22 I can compute and check\n\nSo 1st step is right\n\nlet it be true for some n\n\nwe need to prove for n+1\n\n60*5^n < 6^n\nmultiply both sides by 5\n60*5^(n+1) < 6^n . 5 ...1\n\nbut 6^n. 5 < 6^n . 6 as 6^n > 0\nso 6^n. 5 < 6^(n+1) ....2\n\nfrom 1 and 2\n60.5^(n+1) < 6^(n+1)\n\n\nt QED
921398
I just taught this section today to my calc students.  You start by finding the first and second derivative.  I recommend looking in your book for an example.  Your book probably has something like this towards the front of the section.  It would be too long winded to write everything down here.  If you don't have your book on you, the web page below has problems that you can do and then check your answer.\n\nhttp://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/graphingdirectory/Graphing.html
358699
Please visit http://mathforum.org/library/drmath/view/52240.html
1132233
this question will require calcs using logarithm.\nhence will take time
267038
You might try reading Whitehead and Russell's Principia Mathematica.  I can't promise it will help, as I haven't read it, but I've been told that this issue is addressed.
1181156
It looks like that first 1 should be an i. In this case,\ni^z=exp(i*pi/2*z),\nwhere I am using the principal branch of the logarithm to define the exponential.\nThen the derivative will be \n(i*pi/2)*exp(i*pi/2*z)=\npi/2*i^(z+1)\n\nEasy.
451479
hav u ever heard of the Principle of Mthematical Induction ?\nthat could solve ur problem in a jiffy!!\n\nHowever if u havn't, dont worry!!\n\nNote that, the sum in ur series(i.e. 1+1/2+1/3+...+1/n),where n>=2, is always less than 2, but greater than 1. so it can't b an integer.\nwanna know why?\ntry cramming up some theorems about Sequence & series of real numbers!! its a regulation part of any under-graduate level Real analysis course.
800132
how does this fit into mathematics?
906425
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.
417188
y=Asin(2x)+Bcos(2x)+(x/4)\n\nNow you get to determine why.
914601
http://mathworld.wolfram.com/PoincareConjecture.html
974937
please complete the equation
756805
x<2   (-infinity,2)\nx>3   (3, infinity)\nx<2 see first interval\nx>-2 (-2,infinity)
774953
It's quite complicated.  The link given goes into detail about an inductive method of obtaining this.  If you just want the answer it is -\n\n(n^4 - 6n^3 + 23n^2 -18n + 24) / 24
893324
domain:  from - infinity to + infinity  (that's for x)\n\nrange: 0 to infinity (for y)\n\ny = |x+4|
405166
Try this:\nhttp://www.wtamu.edu/academic/anns/mps/math/mathlab/beg_algebra/beg_alg_tut9_bar.htm
650155
simple\nfirst\nwhen we apply ∑ we can write the series as\n(a_2 - a_1) + (a_3 - a_2) + (a_4 - a_3).... (a_n - a_n-1) (a_n+1 - a_n)\n\nwhen u add \nthe terms starting frm a_2(except a_1 & a_n+1) get cancelled with the correseponding terms and u get\na_n+1 + a_1