Find the partial sum for n=9?\nDoes the series Converge? i know it converges but im messing up somewhere on part a or c or both\nWhat is the sum?
122504
Write 9/[n(n-1)]=9/(n-1)-9/n. The sum is then a telescoping series, so the sum from n=2 to n=9 will be\n9/1-9/2+2/2-9/3....+9/8-9/9=9-1=8.\n\nThe nth partial sum will be 9-9/n, which has a limit\nof 9 as n-->infty. Thus, the sum of the infinite series is 9.
270445
as writen: \n8+1/3, 8+2/3, 8+1/4, 8+2/4, 8+1/5, 8+2/5, 8+1/6, 8+2/6\n\nlooks like it will converge to 8\nif you continue adding 1/x and 2/x where x can be any real number. eventually x will become significantly large compared to 1 and 2 that it will approach 0.\n\nin other words:\nthe limit of 1/x as x -> infinity = 0\n\n\n\nthe second looks like it converges to a limit of 6
372043
1. 101\n2.  There are 2 ones.  The first and last entries of the row.\n3.  Each entry is the sum of the 2 entries above and to     each side of it.\n4.  If the 2'nd entry in a row is a prime, it divides each of the other entries in the row (except for the last entry in the row which is always a 1)\n\nCheck out http://ptri1.tripod.com/ for all manner of kewl sh|t about Pascal's Triangle.  Also Fibonacci Series and other fun things ☺\n\n\nDoug\n\nEDIT:  Give Peter D the points.  He needs 'em worse than I do ☺
409082
Think of a geometric progression.\nHow do you add up 1 + r + r^2 + .. + r^(n-1)? There are n terms, with a common ratio of r, so it is (r^n - 1)/(r-1). So what you want is precisely 1 + x + .. + x^(n-1).\n\nYou just could do a long division as well.\n\nYou could also use this 'trick' (which is used in the proof of the geometric series sum above):\n\nx^n - 1 = (x^n + x^(n-1) + .. + x^2 + x) - (x^(n-1) + .. + x + 1), since all terms cancel out except the first x^n and the last -1.\nBut the first part in brackets is exactly the same as the second, except for a factor of x:\nx^n - 1 = (x^(n-1) + .. + x + 1)(x-1)\nSo (x^n - 1)/(x-1) = x^(n-1) + .. + x + 1.
64187
-9,980,123,110\n\ndisregard -9,980,123,100 from the range -9,980,123,110 to 9,980,123,109\n\nit will become -9,980,123,109 to 9,980,123,109\ntake their sum, it's zero. right?\n\nadd zero to the -9,980,123,110 that we disregarded earlier.\n\nthe answer would be -9,980,123,110
512123
Hmm... 370 and 371 both seem to work.\n\n* * * * * * \n\nI wrote a little program to test all the numbers from 100 to 999.  There are actually four answers:\n\n153\n370\n371\n407\n\nAlso, by "the cube of the sum of the digits", you mean each individual digit is cubed, then you add the sum of the cubes, correct?  For example, 153 is one of the answers because 1^3 + 5^3 + 3^3 = 153.  (It seems like a trivial question if you interpret it the way Dan did, since there are only five numbers to test.)
571284
Please explain what is ~f.  I have curiosity, because neither inverse multiplicative, complex conjugate or inverse function could do that result...\n\nIt should be interesting!! Please post what is ~f !
480054
Let F(x) = Sum n=0toInfty [anx^n] be the generating function of the sequence (an). Written in terms of F, the recurrence relation and the initial condition a0 =1 mean that, for every x (within the radius of convergence) \n\nF(x) = 1 + 8xF(x) + Sum n=0toInfty [10^n x^(n+1)].\n\n(To see this, just use the RR to substitute for an in terms of a(n-1) in the series that defines F; you get the same series but times 8x, plus the other bits.)\nThis gives an equation for F(x) which solves to \n\nF(x) = (1-9x)/[(1-8x)(1-10x)]\n\nExpand this into partial fractions \n\nF(x) = (1/2)[1/(1-8x) + 1/(1-10x)]\n\nand rewrite these as sums of geometric series to get \n\nF(x) = Sum n=0toInfty [(1/2)(10^n + 8^n) x^n]\n\nfor every x, which means that \n\nan = (1/2)(10^n + 8^n)\n\nfor every n.
608172
4(a+b)^2-9(a+b)^2\n\nStep 1. 4(a^2 + b^2) - 9(a^2 + b^2) Distribute the 4 and 9\nStep 2. 4a^2 + 4b^2 - 9a^2 - 9b^2 Combine like terms\nStep 3. -3a^2 - 3b^2 You get the sum\nStep 4. -3(a^2 + b^2) Remove common factors\nStep 5. -3(a + b)(a - b) Factor even more to get final answer (if you multiply this, you should get the sum in step 3)
279724
Add some rigor. Remember that a sequence {a_i} is defind to converge to a limit L iff for every positive difference D > 0, there exists some n such that for all N > n, the difference between the limit L and any subsequent term a_N is less than D. We are squeezing D down to 0 by showing that we can always supply such an n. Thus, your job is to supply some pseudofunction n(D) giving an upper range of indices that satisfy the definition above.\n\nFor the pattern above, the (2n + 1)th term is 8 + 1/(n + 2) while the (2n)th term is 8 + 2/(n + 2). As n increases without bound, the (2n + 1) subsequence approaches 8, as given any positive difference D, one can find a least (2n + 1) such that all terms greater than (2n + 1) yields a difference 8 - (8 + 1/(n + 2)) = 1/(n + 2) less than D. In this case, choose n(D) > (1 - 2D)/D. \nUse a similar tactic to show that the even indexed subsequence converges to the same value, and thus the sequence converges to 8.\n\nAs for the second sequence, as long as all of the remaining values are 6, it converges to 6 by a trivial argument. Given a positive difference D, simply choose n(D) > 6, since 6 - a_n = 0 for any n > 6.
487767
Why not?  If you divide top and bottom of this expression by x^7 you get\n(3 + 4/(x^5))/(9 - 1/(x^4) + 2/(x^7))\n\nEvery term except the 3 and the 9 now has a power of x as denominator, and therefore approaches 0 as x goes to infinity,\ntherefore it approaches 3/9.
329534
psbhowmick, 99 is divisible by 9, but 199 is not. You can't find the remainder simply by looking at the last 2 digits. For instance, the remainder of 22654785/19 is 2, not 9 as you suggest.\n\nDividing by 19, hmmm... the only shortcut I can think of is to use the mod function on your calcutor. Sorry.
156763
Sum of all numbers 1 to 100:\nS = n[2a1 + (n - 1)d]/2\n= 100[2(1) + (100 - 1)]/2\n= 100(101)/2\n= 5050\n\nSum of all numbers 3 to 99, multiples of 3:\nS = n[2a1 + (n - 1)d]/2\n= 33[2(3) + (33 - 1)3]/2\n= 33(6 + 96)/2\n= 1683\n\nSum of all numbers 5 to 100, multiples of 5:\nS = n[2a1 + (n - 1)d]/2\n= 20[2(5) + (20 - 1)5]/2\n= 20(10 + 95)/2\n= 1050\n\nSum of all numbers 15 to 90, multiples of 15:\nS = n[2a1 + (n - 1)d]/2\n= 6[2(15) + (6 - 1)15]/2\n= 6(30 + 75)/2\n= 315\n\nSum of integers from 1 to 100 which are not divisible by 3 and 5:\nS = sum(1-100) - sum(3-99) - sum(5-100) + sum(15-90)\n= 5050 - 1683 - 1050 + 315\n= 2632\n\nWe have to add in the sum(15-90) because in removing sum(3-99) and sum(5-100), we're taking out multiples of 15 twice.  Adding sum(15-90) ensures that we only take out 15 once, 30 once, etc.
483202
2+7\n3+6\n9+0\n\nThat's it. They all add up to 9... I think that's what the question is asking.
481448
This appears to be a geometric series, but I\ndon't understand where the 1 comes from.\nIf a = 100 and ar = 25,\nthen r = 1/4, so the series should run :\n100, 25, 25/4, 25/16, 25/64, 25/256.\nSo I must be wrong.
519525
Hi again Ken,\n\nDid you see my other answer to this similar question?  Did it help any?\n\nIf n is the first of 4 even integers, then the others are:\n\nn+2\nn+4\nn+6\n\nAnd their sum is:\n\nn + n+2 + n+4 + n+6 = 4n + 12\n\nNow, if their sum is 340, then you have the following:\n\n4n + 12 = 340\n4n = 340-12 = 328\nn = 328 ÷ 4 = 82\n\nSo your 4 even integers are:\n\nn = 82\nn+2 = 84\nn+4 = 86\nn+6 = 88\n\nPS - I think the reason you are not getting the answer you want is that the question is worded incorrectly.  Your question states "n is the first 4 even integers" when what I think you mean to say is that "n is the first OF 4 even integers"\n\n~ ♥ ~
582931
Observe:\nsum= 4+12+24+40+60+.....+t(n-1)+t(n)\nsum=       4+12+24+40+.....+t(n-2)+t(n-1)+t(n)\nSubtracting,\nt(n)=4+8+12+16+20+.........upto n terms\n      = n/2[2*4+(n-1)*4]\n      = n/2[4+4n]=n[2+2n]