The terms of the series are defined recursively by the following equations: A1=5   A2=(4n+1)/(5n+3)\n\nDetermine whether the sum of A sub n converges or not
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The series converges because as n increases the ratio gets closer to .8. So A2=9/13 and A3=13/18 and so on. So each iteration makes the +1 in the numerator and the +3 in the denominator more and more negligible.So for n=∞, An=.8 assuming that the series is infinite.
73230
Let's first derive a formula.\nLet\na(1) = first term\na(n) = nth term\nn = number of terms\nd = common difference\ns(n) = sum of the first n terms\n\nThink of a sequence with nth terms.\nNext, add the 1st and nth(or last) terms.\nAdd the 2nd and (n - 1)th(or 2nd to the last) terms.\nAdd the 3rd and (n - 2)th(or 3rd to the last) terms.\n\nNotice something here? They are all equal. Eventually, you will add the nth and 1st terms as the last.\nall of them is equal to n[a(1) + a(n)] Notice that you've added each term twice. That means n[a(1) + a(n)] is twice the sum of the sequence, and the sum is:\ns(n) =  n[a(1) + a(n)]/2\n\nSo, in your sequence,\na(1) = 5\nn = 35\nd = 4\n\nuse the basic arithmetic recursion formula:\na(n) = a(1) + (n - 1)d\na(35) = 5 + (35 - 1)(4)\na(35) = 5 + 34*4\na(35) = 5 + 136\na(35) = 141\n\nwe can now use our derived formula:\n\ns(n) = n[a(1) + a(n)]/2\na(35) = 35[5 + 141]/2\na(35) = 35[146]/2\na(35) = 35(73)\na(35) = 2555\n\nHope this helps^_^
365369
I can give you the answer or give you real help. What is it that you don't understand? E-mail me. And please since I'm being nice enough to answer, be nice enough to chose a best answer.\n\n\nessentially you just have to plug into the formula presented int he wiki entry below. If you know how to do derivatives, there is very little to "understand", unless you have to prove this result true.\nhttp://en.wikipedia.org/wiki/Taylor_series\n\nBasically, you simply need to plug in Taylor's formula, which is based on the derivatives of the numbers below.\n\nYou should know the derivate of ln(5-x) is -1/(5-x).\n(By the chain rule theorem.) The derivate of ln(u) is\n1/u. The derivative of 5-x is -1. So the derivative of\nln(5-x) is 1/(5.-x)\n\nNow the derivative of -1/(u) is 2/(U^2). So the derivative of\n-1/(5-x) is 2/(5-x)^2\n\nEtc\n\nAll you have to do then, is find a value of ln(u) that you know.\nEverybody knows that ln(1) = 0. So chose that.\n5-x = 1 mean x =4.\n\nSo the first co-effcient is 0!!!\n\nNo, for the next, do the same thing using x=4 for the next term.\n1/(5-4) = 1!!!\n\n...\n\nEtc.\n\nLast note:\nA^B means A raised to the exponent \nB.\n\nAs for the radius of convergence, you know that ln(0) goes to infinity. Thus your formula goes to hell once you reach\nx = 5!!! So your radiuus of convergence is 5-4 which is equal to 1!!!
384305
8) The sum of consecutive integers starting with 1 and ending in n is: s=n*(n+1)/2\nThe average of that is s/n\nAnd the effect of removing the mystery number (call it x) from the sum is:\n(s-x)/n=(n+1)/2-(x/n)=31.25\nAs the first term is a multiple of 1/2 in order for this to be true x must be 1/4 or 3/4 of n (it must be less than n remember?).\nTesting 1/4:\n(n+1)/2-1/4=31.25\nn+1=63\nn=62\n\nAs 62 isn't a factor of 4 and we need n to be of the form 4*x (in order for x/n to be 1/4) we can't use this n.\n\nTesting 3/4:\n(n+1)/2-3/4=31.25\n(n+1)=64\nn=63\n\nAs n isn't a factor of 4 there is no solution (n needs to be of the form 4*x/3).\n9) (2^1999)*(5^2001)= (5^2)*(2^1999)*(5^1999)= 25*10^1999\nAs all those 0's aren't going to affect the sum we can remove them. Result: 25\nThe sum of 2 and 5 is 7\n10) x*x + 6x - 91 = (x + 13)(x - 7)\nThe working out is mostly experimentation I'm afraid, but at least the result works. :)\n11) Sorry, no idea how to solve that one.\n12) I'm not sure I understand the question but if it's what I think it is then I can't solve it through non-experimental means. Try looking up combinatorics on wikipedia.
616074
One necessary condition:\n\n1) Suppose the series a0 + a1 + a2 + ..... is absolutely convergent , and\n2) 2^(-n-1) < | bn | < 2^(-n) for all n\n\nThe above two conditions will ensure that the sequence\n\n[(a0/b0) + (a1/b1) + ..... + (an/bn)](bn)\n\nconverges.\n\n[You can easily verify this]
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There are infinite such equations which would produce those numbers, one of which is:\n\nf(n) = n^2 + (π - 36)(n-1)(n-2)(n-3)(n-4)(n-5) / 120\n\nnote that for n = 1,2,3,4,5, the second summand is 0,\nso f(n) = n^2\nbut for n = 6, the second summand is (π - 36)*5*4*3*2*1/120\n= π - 36\nso f(6) = 6^2 + π - 36 = π\n\nderivation of the method of solution is a little complicated.\nsee:\nhttp://en.wikipedia.org/wiki/Newton_polynomial \nor do a search on "newton polynomial"\n\nBasically, if you have a sequence of n numbers:\n\na(1),a(2),a(3), ..., a(n) which are described by the function f(x)\n(in other words, f(1) = a(1), f(2) = a(2), ... f(n) = a(n) )\n\nThen you can rationalize the (n+1)st number to be anything you want (lets say Y) with the function:\n\ng(x) = f(x) + [Y - f(x)] x(x-1)(x-2)...(x-n) / (n+1)!\n\nHere, as like above, if x = 1,2,3,..., n, then the second summand\nis 0, so g(x) = f(x)\n\nbut if x = n+1, then the second summand is \n[Y - f(x)] *(n+1)n*(n-1)*(n-2)*...*3*2*1 / (n+1)! = \nY - f(x)\n\nso g(x) = f(x) + Y - f(x) = Y\n\n(This actually shows that those "what is the next number in this sequence?" questions have infinite answers, since you can create a polynomial which will generate those numbers plus any number you want for the next one!)
534231
I think that the sum converge for all w except w=0. Here is an idea you could try to develop.\nLets define a new serie b(n) like this: b(1) = a(1) + a(2) + ... + a(p1) such that a(1), a(2), ... a(p) have all the same sign and a(p1 + 1) the opposite sign. \nThus b(2) = a(p1 + 1) + .... a(p2) such that a(p1 + 1), ... a(p2) have the same sign and a(p2 + 1) the opposite sign.\nThus b(3) = ....  etc   Note that this construction is possible only if w is different from 0.\nThen, by construction the b serie is an alternating one. It is easy to verify that |b(n)| is a serie which converges to zero. The problem would be ended if we had |b(n+1)|<|b(n)|, but that is not true. Nonetheless, you can make an integral approach to find a majoration of the |b(n)|. [something like a majoration of |integral of cos(wx)/x between pi*n/w and pi*(n+1)/w |]\nI leave you the detail of calculation... this approach should work ...
1079
What you are describing is an infinite series using a decreasing geometric progression. The sum of such a series can be described by this proof:\n\nLet T = total distance traveled (in spaces)\nThen T = 1 + (1/2) + (1/4) + (1/8) + (1/16) ...\nAlso T - 1 = (1/2) + (1/4) + (1/8) + (1/16) ...\nAnd if we multiply both sides of the equation by 2:\n2*(T - 1) = 2*((1/2) + (1/4) + (1/8) + (1/16) ...)\nMultiplying through:\n2T - 2 = 1 + (1/2) + (1/4) + (1/8) + (1/16) ...\nBut the Right side is the same as T!\n2T - 2 = T\nSimplifying:\nT = 2\nSo the total distance traveled is 2 spaces.
245841
Hi! This recurrence relation can be evaluated for n=5^s, s>0, since we have been given T(1).\n\nUsing the given formula,\n\nT(625)=7*T(125) + 10*625\n= 7*[ 7*T(25) + 10*125 ] + 10*625\n= 7*[ 7*{ 7*T(5) + 10*25 } + 10*125 ] +10*625\n= 7*[ 7*{ 7*( 7*T(1) + 10*5 ) + 10*25 } + 10*125 ] +10*625\n\nNow, T(1)=1\n\nInstead of evaluating the expression, open the brackets, to get,\n\nT(625) = 7^4 + 7^3*10*5 + 7^2*10*25 + 7^1*10*125 + 10*625\n\nwhich can also be written as:\n\nT(625) = (7^4) + (7^3)*10*(5^1) + (7^2)*10*(5^2) + (7^1)*10*(5^3) + (7^0)*10*(5^4)\n\nSo, in general for n=5^s, we can also express the relation as,\n\nT(5^s) = 7^s + Σ{7^(s-k-1)*10*5^(k+1)} where k varies from 0 to (s-1), provided s>0.\n\nThe value of T(625) from the expression above comes out to be: 46801
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1. The equation for the first would be                      \n\nt(n)={2^(5-1)}*{3^(n-1)}   It is a A.P.\n\n2. It is neither
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18. Use induction.  If n=0, then it's just (f*g)(a) = f(a)g(a).\n\nAssume that Leibniz' formula holds for some n>=0, and show that it holds for n+1.\n\n(f*g)^(n)(x) = sum k=0 to n (n choose k) f^(k)(x) g^(n-k)(x)\n\nDifferentiate this with respect to x:\n\n(f*g)^(n+1)(x) = sum k=0 to n (n choose k) [ f^(k+1)(x) g^(n-k)(x) + f^(k) g^(n-k+1)(x) ]\n= sum k=0 to n (n choose k) f^(k)(x) g^(n+1-k)(x) +\nsum k=1 to n+1 (n choose k-1) f^(k)(x) g^(n+1-k)(x)\n= f^(0)(x) g^(n+1)(x) +\nsum k=1 to n [(n choose k)+(n choose k-1)] f^(k)(x) g^(n+1-k)(x)\n+ f^(n+1)(x) g^(0)(x)\n\n(n+1 choose 0) = (n+1 choose n+1) = 1, and\n(n choose k) + (n choose k-1) = n!/[k!(n-k)!] + n!/[(k-1)!(n-k+1)!]\n= [(n-k+1)n! + kn!]/[k!(n+1-k)!] = (n+1)!/[k!(n+1-k)!] =\n(n+1 choose k)\n\nputting all the terms together gives the result.\n\n22a.Suppose a is a double root of f.  Then f(x)=(x-a)^2*g(x), so a is a root of f.  By the product rule, f'(x)=g(x)*2(x-a) + (x-a)^2*g'(x), so f'(a)=0.  Suppose f(a)=0 and f'(a)=0.  Then f(x) = (x-a)r(x) for some polynomial r(x).  f'(x) = r(x) + (x-a)r'(x), and since f'(a)=0, it follows that r(a)=0, so r(x)=(x-a)g(x), meaning f(x)=(x-a)^2*g(x).\nb. f'(x) = 2ax + b, so its root is -b/2a.  For this to be a root of f(x) as well, must have a(-b/2a)^2 + b(-b/2a) + c = 0, or b^2 - 4ac = 0.  Geometrically, the graph is tangent to the x-axis at x=-b/2a.\n\n27. Let L = f'(0), and define g(x) = f(x)/x for nonzero x, and g(0) = L.  Then g(0) = L = f'(0) = lim (x->0) f(x)/x = lim (x->0) g(x), so by the definition of continuity, g is continuous at 0.
647787
(n/(n - 3)) + (5/(2n - 6))\n(n/(n - 3)) + (5/(2(n - 3)))\nMultiply everything by 2(n - 3)\n(2n + 5)/(2n - 6)
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I cannot prove it formally but I can mention the steps\n\na | u(n)\n\na | u(n+1)\n\nso a | u(n+1) - u(n) if a devided b and a devided c then it devides b-c\n\nso a | u(n-1)\n\nthere by we reach a | u(0) and a|u(1)\n\nu(0) does not give any infomation as u(0) =0\n\nso a | u(1) = 1\nso a= +/-1 \nu(0) = 0\nu(1)  = 1\n\nnext terms are u(2) = 1,u(3) = 2 ,u(4) = 3, then 5,8,13,
481707
You're answer is perfectly correct.  You do want to use root test, and you applied it properly.\n\nYou could also check your work by sorta "thinking" it out.\n\nthe first term (if you start with n=1) is 1, then 1/4 then 1/27 then 1/256 and so on, so as you can see, if you sum the first 3 terms we get a number close to 1.3, if we sum the first 4 numbers its still about 1.3... doesn't matter how many terms it is, i think it ends up around 1.3 or 1.33333 or something, there is a way to figure out what the sum converges to.
422933
the 1st three terms means n=1, n=2, n=3 so substitute\n\nThis gives, 3*1+2,3*2+2,3*3+2\n\n=5,8,11
866
C = A - B = infinity - infinity = undefined :-), not 0.\n\nIf you have the specific series for A and B, then you could technically substitute them into C and figure out C for different {A,B}.\n\nThe short answer is this: not all infinity series are the same, so their difference is really not zero.
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The flaw in that sort of reasoning is due to the untrue assumption that the solution of all infinite series is infinite.  For example, the sum of the series \n\n1/2 + 1/4 + 1/8...\n\nhas infinite terms, but a finite solution of 1.  Applied to time and distance, your problems are solved.\n\nGary H