Use induction to prove that for all integers n >=1,  1 x 1 ! + 2 x 2 ! + - - - n x n ! = ( n + l ) ! - 1?
958370
We wish to show that:\n[k=1, n]∑kk! = (n+1)! - 1\n\nClearly, this is true for n=1, since 1*1! = 2! - 1 = 1. Suppose it is true for some n. Then for n+1:\n[k=1, n+1]∑kk!\n(n+1)*(n+1)! + [k=1, n]∑kk!\n(n+1)*(n+1)! + (n+1)! - 1\n(n+2)*(n+1)! - 1\n(n+2)! - 1\n\nWhich is what we would get if we substituted n+1 for n in our formula. So by induction, the formula holds for all n. Q.E.D.
191250
Mathematicians have proven that, in general, 1 + 1 = 2. It took them awhile to do it, but they succeeded. Unless the time truly comes when there is a need to ignore this generalisation, this assumption made is as good as any other axioms ever declared by the wise logicians.
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)]
151956
Exponentiation is usually taken to bind more strongly than addition, so 10^n+1 is usually interpreted as (10^n)+1.  You intended 10^(n+1).\n\nLet Rep(7, 10, n) be the "repunit" the integer the base 10 representation of which is n seven's.  You wish to prove by Mathematical Induction that the statement P(n) defined as\n\ndefine: P(n) states Sum{i = 1 to n}{Rep(7, 10, i)} = (7/81)*[10^(n+1) - (9*n) - 10]\n\nholds for all positive integers n = 1, 2, 3, ....\n\n\nFirst prove the lemma that Rep(7, 10, i) = "i seven's in base ten" = Sum{k = 0 to i-1}{7 * (10^k)} = 7 * ((10^i - 1) / 9) for positive integer i = 1, 2, 3, ... either by the formula for the sum of a geometric progression or by Mathematical Induction. :-)\n\n\nNow consider the statement P(1).  The left hand side of the equality is Sum{i = 1 to 1}{Rep(7, 10, i)} = Rep(7, 10, i) = 7.  The right hand side of the equality is (7/81)*[10^(1+1) - (9*1) - 10] = (7/81)*(100 - 9 - 10) = (7/81)*81 = 7.  The left hand side equals the right hand side so P(1) is true.\n\nNow let m be a positive integer m >= 1 and assume the statement P(m) is true with the goal of proving that the P(m+1) must then also be true.  For the statement P(m+1) the left hand side of the equality is\n\nSum{i = 1 to m+1}{Rep(7, 10, i)} = Sum{i = 1 to m}{Rep(7, 10, i)}  +  Rep(7, 10, m+1)\n= Sum{i = 1 to m}{Rep(7, 10, i)}  +  7 * ((10^(m+1) - 1) / 9)      # by the lemma\n= (7/81)*[10^(m+1) - (9*m) - 10]  +  7 * ((10^(m+1) - 1) / 9)      # by the assumption of P(m)\n= (7/81)*[10^(m+1) - (9*m) - 10]  +  7 * (9 * (10^(m+1) - 1) / 81) # multiply both numerator and denominator of right addend by 9\n= (7/81)*[ (10^(m+1) - (9*m) - 10)  +  (9 * (10^(m+1) - 1)) ]      # "undistribute" common factor of 7/81\n= (7/81)*[ (10^(m+1) - (9*m) - 10)  +  (9 * 10^(m+1) - 9) ]        # distribute multiplication by 9\n= (7/81)*[ ( 10^(m+1) + 9*10^(m+1) ) - 9*m - 9 - 10 ]              # addition is commutative and associative\n= (7/81)*[ 10 * 10^(m+1) - 9*(m+1) - 10 ]                          # more distributuve law\n= (7/81)*[ 10^((m+1)+1) - 9*(m+1) - 10 ]                           # definition of exponentiation\n= right hand side of the equality of P(m+1)\n\nThus both P(1) is true and "for all integers m >= 1, P(m) implies P(m+1)" is true.  You can knock over the first domino and each domino when it falls will knock over the next.  So by Mathematical Induction "for all integers n >= 1, P(n)" is true and all of the dominoes eventually fall.
661659
(0! + 0! + 0! + 0!)!\n=(1+1+1+1)!\n=(4)!\n=24
258926
Sometimes it helps to look at a pattern.  Here,\n5! = 5x4x3x2x1 = 120\n4! = 4x3x2x1 = 24, which is 5!/5\n3!= 3x2x1 = 6 which is 4!/4\nso 2!= 3!/3\nand 1!= 2!/2\n\nso 0! = 1!/1 = 1/1 = 1
806805
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.
970176
Not without a fallacious proof like the one above.  (Notice how the "proof" includes a division by (a-b) which is a division by zero.  This is not allowed and leads to the seeming contradiction that 0 = 1.\n\nZero does *not* equal One.\n\nYou could use similar fallacious methods:\n\n1^0 = 1\n1^1 = 1\n\nTake the log base 1 of both sides:\n0 = log 1 (1)\n1 = log 1 (1)\n\nTherefore 0 = 1.
720161
Yes, I can:\n4! + 4 \n=\n(4 x 3 x 2 x 1) + 4 = 28\n\n(BTW, 4squared + 4squared = 32)
833237
so far, so good\n\nn!/(n-4)! = 56 [n!/(n-2)!] \n\nmultiply both sides by (n-2)!/n!\n\n[n!(n-2)!] / [n!(n-4)!] = 56\n\n(n-2)! / (n-4)! = 56\n\n(n-2)(n-3) = 56\n\nYou could multiply this out and solve the quadratic, in which case you will get one positive and one negative root, or if you remember that 56 = 8*7, then that quickly leads to n = 10
269006
Ask your friends if they can find the error in the following proof.\n(it's obviously not right, but it looks cool!)\n\na=b\na+1 = b+1\n(a+1)^2 = (b+1)^2\na^2 + 2a +1 = b^2 + 2b +1\ncancel the ones and rearrange\na^2 - b^2 = 2(b-a)\n(a+b)(a-b) = 2(b-a)\ntherefore a+b = -2\nPick numbers for a and b\nand look!! 5 + 7 = -2\nThus shown.\nLOL
972511
A fascinating subject.\n\nIt's recursively defined.  \n\nf(n+2) = f(n+1) + f(n) for n>0\n\nwhere f(1) = f(2) = 1.\n\nIt has some really neat properties and shows up in nature.
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)
155864
yes, it equals x(ln(x)-1)
277221
Here is a proof by induction on p, which is probably what you are looking for.\n\nNote:\nTo label equations that I will refer to later, I will use a number in parentheses followed by a colon just before the equation, for example "(4): x = 0". To refer back to a labelled equation, I will use the equation number in parentheses, for example "by (4), blah blah".\n\n11^p - 4^p is divisible by 7 for any integer p>=1.\n\nProof:\n\nBase step:\n-----------------------\nIf p=1, then\n\n11^p - 4^p\n= 11 - 4\n= 7,\n\nso 7 divides 11^p - 4^p when p=1.\n\nInductive step:\n-----------------------\nFix some p>=1 and suppose the desired result holds for p, i.e. that 11^p - 4^p is divisible by 7. Using this, we want to show that 11^(p+1) - 4^(p+1) is divisible by 7.\n\nFirst, since we assumed 11^p - 4^p is divisible by 7, we know that\n\n(1): 11^p - 4^p = 7*k\n\nfor some integer k. Then\n\n11^(p+1) - 4^(p+1)\n= 11*11^p - 4*4^p\n= 7*11^p + 4*11^p - 4*4^p\n= 7*11^p + 4*(11^p - 4^p),\nnow substituting from (1),\n= 7*11^p + 4*7*k\n= 7*(11^p + 4*k).\n\nSo 11^(p+1) - 4^(p+1) is divisible by 7.\n\nTherefore by the principle of mathematical induction, 11^p - 4^p is divisible by 7 for every integer p>=1.\n\nEnd of proof.\n\nHope this helps :)
95089
1(1)+0(2)+1(4)+1(8)+1(16)+0(32)+0(64)+0(128)+1(256)=\n1+0+4+8+16+0+0+0+256=\n285
674498
1+1=2. That's it. The answer for everything else would be simply 42