Prove (1^3)+(2^3)+...+(n/(9n+1)!) = 1-(1/(n+1)!)
137899
It's so obvious...I'll let you do your own homework.
482299
I think you meant 3^0 + 3^1 + 3^2 + ... + 3^(n-1)\nNote the parenthese.\nIt reduces to (3^n - 1) / 2\n\na^0 + a^1 + a^2 + ... + a^(n-1) reduces to (a^n - 1) / (a - 1)
506931
You can easily solve this by realizing (4m - 3n)(4m + 3n) has a pattern (a + b)(a - b), which equals to a^2 - b^2\n\nSo,\n(4m - 3n)(4m + 3n) = (4m)^2 - (3n)^2\n= 16m^2 - 9n^2
441735
You have:\n\n2(-3^1 - 1) + 2(-3^1 - 1) + 2(-3^1 - 1) + 2(-3^1 - 1) + 2(-3^1 - 1) + 2(-3^1 - 1)\n\n3^1 = 3\n3^2 = 9\n3^3 = 27\n3^4 = 81\n3^5 = 243\n3^6 = 729\n\n2(-3 - 1) + 2(-9 - 1) + 2(-27 - 1) + 2(-81 - 1) + 2(-243 - 1) + 2(-729 - 1)\n\nOk....I assume you can add and subtract and all that stuff. You can take it from here.
474840
we need to prove this by induction...\n\nfor n belonging to Z\n\nf(n)=n*(n+1)*(2n+1)/6\n\nfor n=1\n\nLHS:\n f(1)= 1^2 =1\nRHS\n1*(1+1)8(2*1+1)/6=6/6=1\n\nThus its true when n=1;\n\nAssume f(k) is true i.e. \n1^2+2^2+......+k^2=k(k+1)(2k+1).. Let this be equation 1\n\nNow we need to prove f(k+1) is true...\nie\nf(k+1)=(k+1)(k+2)(2(k+1)+1)/6\n          =(k+1)(k+2)(2k+3)/6\nL.h.s\n\nf(k+1)=1^2+2^2........+k^2+(k+1)^2\n          =k(k+1)(2K+1)/6 + (k+1)^2 from eq 1\n          =(k+1)(k(2k+1)/6 +k+1)\n          =(1/6)(k+1)(2k^2+k+6k+6)\n          =(1/6)(k+1)(2k^2+7k+6)\n          =(1/6)(k+1)(k+2)(2k+3)\n\nHence f(k+1) is true if f(k) is true....\n\nThus its true for all positive integers.......
292904
Nice problem. Here's the solution: 1000! is a multiple of 9 which is the same thing with saying that: S(1) = the sum of its digits is a multiple of 9 (Using the following known fact: a number is divisible by 9 if and only if the sum of its digits is divisible by 9). \n   Clearly S(1) < 1000! and 9 | S(1). Summing the digits of S(1) one obtains a number S(2) which is strictly smaller than S(1) and is a multiple of 9. And so on, until finally one arrives at a one digit number, say S(r) for some r  which is divisible by 9. But this can only be 0 or 9. Since S(r) is the sum of the digits of the previous number obtained by this algorithm, namely S(r-1), it can not be 0. So it must be 9! \n   Note that this type of proof works for any number divisible by 9, not only factorials of numbers, and in fact, with exactly the arguments displayed above it can be prooved that the single digit after adding the digits of an arbitrarly chosen number N is the digit d > 0 such that N-d is a number divisible by 9. (If you don't belive it, simply work out a computer programme that verifies this for all the numbers smaller than 1 000 000 for example).
367048
(4)/(x+7) + (x+3)/(x+7)\n=(4 + (x+3))/(x+7)\n=(4 + x + 3)/(x+7)\n=( x +7)/(x+7)\n=1
421526
You can prove it by induction.\n\nIt is obviously true when n = 1.\n\nInduction pypothesis:  Suppose it is true for n\n\nWe need to show  is true for (n+1).\n\nDoes 24 divide (5^(2*(n+1))-1))?\n\nWe can rewrrite this as:\n\n25*(5^2n)-1  Subtract 24 from this to get 25*(5^2n)-25 \n\nThis is just equal to 25*(5^2n)-1).\n\nThis has to be divisible by 24 -- since one of the factors is (5^2n)-1) and our induction hypothesis says that it is.  \n\nAdd 24 to it and we get something that is also divisible by 24 -- but it is (5^(2*(n+1))-1)) -- so our induction is done.
483066
3|(2x-1)/5| + 3 = 1?\nThats impossible. |anything| >= 0, so the LHS is >= 3. Did you mean something else?
397271
LetP(n) denote the statement\n1^2+2^2+3^2+......+n^2=n(n+1)(2n+1)/6\nwhen n=1,P(1)=1(2)(3)/6=1 checks\nlet P(k) be true\ni.e.1^2+2^2+3^2+.......+k^2=k(k+1)(2k+1)/6\nto prove\nP(k+1)=(k+1)(k+2)(2k+3)/6 is true\nP(k+1)=1^2+2^2+3^2+.......+((k+1)^2\n=>[1^2+2^2+3^2+......+k^2]+(k+1)^2\nnow 1^2+2^2+3^2+......+k^2=k(k+1)(2k+1)/6\nsubsttituting\n=k(k+1)(2k+1)/6 +(k+1)^2\n=(k+1)[k(2k+1)+6(k+1)]/6\n=(k+1)(2k^2+7k+6)/6\n=(k+1)(k+2)(2k+3)/6\nhence proved\n\n2.try by your self
32840
There are a couple of ways to work this out, I will take the easiest to explain.\n\n(x-3)(x^3+3x^2+9x+27) Can be rewritten as:\n\n(x)(x^3+3x^2+9x+27)+(-3)(x^3+3x^2+9x+27) Now, multiply those out.\n\nx^4+3x^3+9x^2+27x-3x^3-9x^2-27x-81 Now, sort by like powers of x\n\nx^4+3x^3-3x^3+9x^2-9x^2+27x-27x-81 Simplify\n\nx^4-81
1932
(n+k-1) over (k) = (n+k-1) / (k)\n\n=> (n+k-1) / (k) = n/k +1 -1/k
296534
1st step: try for n = 1\n\nThen 1*2^1 = 2 and\n(1-1)*2^(1+1)+2 = 2\n\n2nd step: Assume true for n, show true for n+1.\n\nThen sum(i=1 to n) = (n-1)*2^(n+1)+2\nThen for n+1:\n\n(n-1)*2^(n+1)+2 + (n+1)2^(n+1)\n= (n-1 + n+1)2^(n+1)+2\n= 2n2^(n+1)+2\n= n2^(n+2)+2\n= ((n+1) - 1)*2^((n+1)+1) + 2\n\nThus it works.
14593
(x-2)/(x^2-2x-15)+(x+2)/(5-x)\n(x-2)/((x+3)(x-5))+(x+2)/(5-x)\n(x-2)/((x+3)(x-5))+(-2-x)(x+3)/(x-5)(x+3)\n\n[(x-2) + (x+3)(-2-x)] / [(x+3)(x-5)]\n\nANSWER :\n(x^2+4x+8)/((x+3)(5-x))
226024
Let m = n = 1.\nThen m = n.\nThen m^2 = mn.\nThen m^2 - n^2 = mn - n^2.\nThen (m+n)(m-n) = n(m-n).\nThem m+n = n.\nSo 1+1=1.\n\nSo the answer is 1.\n\nedit: raja, don't make me come over there!
168206
!!, !!!!!!!! are "multifactorials". They are described in the first attached article.\n\nBriefly,\nn!! = 1 for n=0 or 1,\nn!! = n * (n-2)!! for n>=2.\n\nn!!...! (k "!" signes) = 1, for n<k\nn!!...! (k "!" signes) = n * (n-k)!!...! (k "!" signes)\n\n(2n)!! = n! * 2^n\n(kn)!!...! (k "!" signes) = n! * k^n\n\nSo, we have\n1) 6! = 1*2*3*4*5*6 = 720\n2) 6!! = 2*4*6 = 48\n3) 0! = 1\n4) 5!! = 1*3*5 = 15\n(5!!)!! = 15!! = 1*3*5*7*9*11*13*15 = 2027025\n\nOther factorials are too huge to compute precisely, but we can evaluate them using Stirling formula (second article)\nln n! ~= n*(ln n) - n + (ln(2*pi*n))/2 + 1/(12*n)\n\n5) 5! = 120\n(5!)! = 120! =~ e^(120*ln(120) - 120 + (ln(2*pi*120))/2 + 1/(12*120)) =~ e^457,81 =~ 6.69*10^198\n6) 1234!!!!!!!! = 1234 * 1233 * 1232!!!!!!!! =\n= 1521522 * (154*8)!!!!!!!! =\n= 1521522 * 154! * 8^154 = e^972.43 =~ 2.1*10^422\n\nThus, 6!! - 6! + (5!)! - 1234!!!!!!!! - 0! + (5!!)!! =\n=~ 48 - 720 + 6.69*10^198 - 2.1*10^422 - 1 + 2027025 =~ -2.1*10^422\n\nAnswer: about -2.1*10^422
418160
(n + 3)(2n + 3) = (n + 2)^2 + (n 2)2^2\nFirst use FOIL to simplify\n2n^2 + 3n + 6n + 9 = n^2 + 2n + 2n + 4 + n^4\nCombine like terms on each side\n2n^2 + 9n + 9 = n^4 + n^2 + 4n + 4\nMove everything over to one side.\nn^4 - n^2 - 5n - 5 = 0\nFactor.\n(x+1)(x^3 -x^2 -5) = 0\nThat means x+1 = 0 or x^3 -x^2 -5 = 0\nx+1 = 0 when x = -1\nx^3 -x^2 -5 = 0 when  = 2.1163