Can Anyone tell me how to prove this Equation by Mathematical Induction ?
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.
574891
im pretty sure that you will have to use the parts method to solve it. but the answer is\n\nx(log(1-ie^(ix))-log(1+ie^(ix))+i(Li(-ie^(ix)-Li(ie^(ix))\n\ni=imaginary unit\nlog=natural logarithm\nLi=polylogarithm\n\nyou can check my answer to see if its right because i also thought that it was wrong. i dont know how the imaginary units could have shown up but o well.
131346
Prove that it really is possible! I really would like to see it.\n\nAssume x = y\n\nThen\n\nx + y = y + y  (Add y to both sides of the equation)\n\nx - y + y = y  (Subtract y from both sides)\n\n(x-y+y)/(x-y) = y/(x-y)\n\nSimple Regrouping:\n(x-y)/(x-y) + y/(x-y) = y/(x-y)\n\nWe all see that (x-y)/(x-y) = 1 we can also\nAdd 1 + 1 to both sides \n\n1 + y/(x-y) + 1 + 1  = y/(x-y) + 1 + 1\n\nHey, actually bot sides have a common term of y/(x-y) so let's just get rid of them by subtract them from both sides\n\n1 + 2 = 1 + 1\n\nHey, what do you know.\n3 = 1 + 1\n\nOk what did I do wrong guys. :p
67383
1. A(N)<2 for any N:\n\na) A(0) = sqrt(2) < 2\nb) induction: A(N)<2 => A(N+1) = sqrt(2)^A(N) < sqrt(2)^2 = 2\n\n2. The sequence A(N) increases:\n\na) 1<sqrt(2) => sqrt(2)^1<sqrt(2)^sqrt(2) => A(0)<A(1)\nb) induction: A(N)<A(N+1) => sqrt(2)^A(N)<sqrt(2)^A(N+1) => A(N+1)<A(N+2)\n\nFrom 1 and 2 it follows that A(N) HAS a limit X, with sqrt(2)<X<=2\n\n3. sqrt(2)^X=X:\nX = lim(N->inf.)A(N) = lim(N->inf.)A(N+1) = lim(N->inf.)sqrt(2)^A(N) = sqrt(2)^X\n\n4. X=2 or X=4:\nWe already know that 2 and 4 are solutions for sqrt(2)^X=X. Let's show that they are the ONLY solutions:\nThe second derivative of f(X)=sqrt(2)^X-X is f"(X) = ln(sqrt(2))^2 * sqrt(2)^X > 0 for any X. Then f is a strictly convex function, so f(X)=0 can have AT MOST two solutions. We already found two, so they must be the only ones.\n\n5. X=2:\nWe showed that X<=2 and (X=2 or X=4), you draw the conclusion.
848097
I'll haev a go at number 2.\n\na) So we prove that sum(k=1 to n)k + sum(k=1 to n-1) = n^2.\nSince the sum of the first n natural numbers is n(n+1)/2, this is equal to\nn(n+1)/2 + (n-1)n/2\n= (n^2 + n + n^2 -n)/2\n= 2n^2/2\n= n^2.  QED.\n\nb) We prove that (x^0 + x^1 + x^2 + ... + x^n)^2 = x^0 + 2x +... + (n+1)x^n + nx^(n+1) + ... + 2x^(2n-1) + x^(2n).\nThe LHS will generate n^2 terms x^k where k is an integer.  How many identical terms x^k will there be for a given k?\n\nCase 1) If k<=n, there are no restrictions on k so there are k+1 multiplications resulting in x^k which are x^0*x^k, x*x^(k-1), x^2*x^(k-2), ... ,x^k*1.  The coefficient is k+1.\nCase 2) If n<k<=2n, the products resulting in x^k are x^n*x(k-n), x^(n-1)*x^(k-n+1), ... x^(k-n+1)*x^(n-1), x^(k-n)*x^n, a total of (2n-k+1) terms.  The coefficient is 2n-k+1.\n\nSo the coefficient is 1 for x^0, increases by 1 up to (n+1)x^n, and then decreases by one up to x^2n.\n\nSetting x=10 demonstrates 1*1=1, 11*11=121, etc.\n\n\nc) n^2-n is divisible by 2 for all n.  Proof: if n is even then n^2 is even and n^2-n is also even.  If n is odd then n^2 is odd and n^2-n is odd.\n\nNow n^3-n = n(n^2-1) = n(n+1)(n-1).  This is the product of 3 consecutive numbers, one of which must be a multiple of 3.\n\nThe property doesn't hold for n^4-n, for example 2^4-2 = 14 which isn't divisible by 4.
971390
{3XY^2M/1}-{2M/[X^-1Y^-1]^2\n+{[4X^2Y^2]/M^-1}-\n(3X^(3Y)^(3M)^-1/(XYM)^-2
1004368
x^4+y^4+8>= 8xy? \n\nI redid the proof using the Am-Gm inequality.\nUsing Am-Gm\nlet x^4=(x²)² & y^4=(y²)²\n((x²)²+ (y²)²)/2 >= sqrt{(x²*y²)²} =>\n\n((x²)²+ (y²)²)>= 2*x²*y²\n\n((x²)²+ (y²)²)+8>= 2*x²*y²+8\nUsing Am-Gm again on 2*x²*y²+8 =>\n(2*x²*y²+8)/2>=sqrt(2*x²*y²*8) =>\n(2*x²*y²+8)>= 2*sqrt(16*x²*y²) =>\n(2*x²*y²+8)>= 2*4*x*y=8xy\nSince x^4+y^4+8>= (2*x²*y²+8) &\n(2*x²*y²+8)>=8*x*y\nthen x^4+y^4+8>= 8xy. \nend of proof using the Am-Gm twice
134713
It can be proved with the use of the integral, but you don't have to.\n\n1)The first some is n(n+1)/2\n\nTo prove that you can add all the consecutive numbers twice, but rearraging them, like that: 1+n; 2+(n-1); 3+(n-3)\n...n+1. Then you have n times the number n+1, which means that this sum equals n(n+1). But since you added twice the sum you wanted, to get the desired result, divide by 2.\nThus you obtain n(n+1)/2\n2)Prove (with the use of mathematical induction) that the sum of all cubes of consecutive integers from 1 to n is  \n[n^2*(n+1)^2]/4\n3)Elevate the first sum, n(n+1)/2, to the second power as requested. See that you get \n[n(n+1)/2]^2= [n^2*(n+1)^2]/4\n4)Observe that the results of 2) and 3) are the same.\n\nQ.E.D\n\nGood Luck.
284229
First you have to arrange it so the powers of x count down, which they already do.  I think you have to also get the leading coefficient = 1, but I'm not sure - better check on that.  It is a second-degree trinomial b/c there are 3 terms (tri=3, trinomial) and the highest power of x is 2 (2nd degree).
790751
you have to do it by induction:\nbasically:\nn=2:\nka_1 + ka_2 = k( a_1 +a_2)\n\nnow, assume it is true for n\n\nlet us prove it for n+1:\n\nsum_{i=1} ^{n+1} ka_i = sum_{i=1} ^{n} ka_i + ka_{n+1}\n= k ( sum_{i=1} ^{n} a_i )+ ka_{n+1} now you have only two summands, so:\n= k( sum_{i=1} ^{n} a_i  + a_{n+1} )\n= k sum_{i=1} ^{n+1}  a_i \n~
304206
the clue would be to simplify the equation  to \nlim {x +[-(x+1)+1]}/(x+1)\n x-->0 \nlim [x + 1 -x]/(x+1)\nx-->\nlim (1/(x+1)) =1\nx-->0
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
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.
850884
using chain rule and product rule\n\ny'= 3x^(2)*tan(4x)\n         +4x^(3)*sec^(2)(4x)\n  = x^2(3tan(4x)+4x*sec^(2)(4x) \n\ni hope that this helps
300657
let a = b  \nMultiply both sides by a \na² = ab  \n Add (a² - 2ab) to both sides \na² + a² - 2ab = ab + a² - 2ab  \n Factor the left, and collect like terms on the right \n2(a² - ab) = a² - ab  \n Divide both sides by (a² - ab)\n2 = 1\n(The place where the proof goes wrong is the very last step.  If a² - ab = 0, then you would be dividing by zero, which is undefined.)
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
1033116
A X = D\nX = A^(-1) D. (A inverse * D)\n Do it ur self.