What would the sum of This pattern equation be(i know the answer, first to get it right gets the 10 points):\n1/1+2 + 1/2+3 + 1/3+4..... + 1/99+100\n\nChoices: A. 100/101?  B. 1?   C.99/100?   C.1000?
210462
Answer is B. I reexpressed your question to summation of 1 to 99 with this expression 1/(2n+1).\n\nNext divide top n bottom by n which gives (1/n)/(2 + 1/n)\nThen using L.Hopital rule differentiate top and bottom with respect to n gives (-1/n^2)/(-1/n^2) = 1
402227
The you work it out thus:\n\n0.1 = 10 minus 1\n0.01 = 10 minus 2\n0.001 = 10 minus 3\n1=\n10 = 10 plus 1\n100 = 10 plus 2\n1000 =10 plus 3
287293
this is possible to figure out.  What it is a combination of numbers   i will do examples to keep it simple.   If your number was 00  or 000  then the answer was 9.  well its a pattern    if your number is 001 then the answer he came up with was 8.  well then if you had 002  then it was 7.  but also on the note if your number was 011 or some combination of this number then it was 8 also  because when you take 011 and add 0+1+1 you get 2 so it was 8.   the way it works is the numbers you type in it adds them together so there is a correlating number that goes with the answer that was added together.  so... if the numer added together was 0 then it was 9.  and the number added together was 1 then he came up with 8.  added together you got 2 then he said 7.  and just keep going ... 3 is 6,   4 is 5.    also on the same account    5 is 4,  6 is 3,   7 is 2,   8 is 1,  and 9 is 0.
121662
The first term is 7C0 d^7 (2e)^0\nThe second is 7C1 d^6 (2e)^1\nThe third is 7C2 d^5 (2e)^2\nThe fourth is 7C3 d^4 (2e)^3\nSee how the pattern works -\n\nThe fifth term is 7C4 d^3 (2e)^4\n\n7C4 = 35; (2e)^4 = 16e^4; 35 times 16 = 560\n\nSo it simplifies to 560 d^3 e^4.
404412
The summation formula you used before,\n\n4n(n+1)/2 - 5n,\n\nsums 4x-5 from 1 to 33, giving you 2079.\n\nYou want to sum from 4 to 33, so you need to subtract off the terms corresponding to 1,2,3.  That is, you want\n\n4n(n+1)/2 - 5n - [ 4k(k+1)/2 - 5k ]\n\nwhere n=33, and k=3.
394271
This is solvable, and you wrote it very well!\n\nThe key to setting this up is knowing that for any 2-digit number, you can write it as 10a + b.\n\nSo your problem states that a+b=9 (the sum of the tens digit and ones digit is 9), and it states that a=3b+1 (the tens digit is 1 more than 3 times the ones digit).\n\nSo now you have 2 equations in 2 unknowns, which you can easily solve for:\n\na + b = 9\na = 3b + 1\n\nSubstitute a = 3b + 1 into the first equation to get:\n\na + b = 9\n(3b + 1) + b = 9\n4b + 1 = 9\n4b = 8\nb = 2\n\nNow you can easily solve for a, which equals 7 since a+b=9.\n\nSo your 2-digit number is 72.\n\nCheck: 7+2=9\n7 = 1 + 3*2\n\nIt checks out, so our answer is correct!\n\n~ ♥ ~
317366
try seeing OEIS: http://www.research.att.com/~njas/sequences/?q=1%2C+1%2C+3%2C+5%2C+11%2C+21%2C+43&sort=0&fmt=0&language=english&go=Search\n\neverytime you had an integer sequence problem, you could check there, most of the time, they had an entry of the possible sequence there.\n\nthe most possible answer to your question is perhaps the \nJacobsthal sequence: a(n) = a(n-1) + 2a(n-2). \n\nwhich gives the sequence\n0, 1, 1, 3, 5, 11, 21, 43, 85, 171, 341, 683, 1365, 2731, 5461, 10923, 21845, 43691, 87381, 174763, 349525, 699051, 1398101, 2796203, 5592405, 11184811, 22369621, 44739243, 89478485, 178956971, 357913941, 715827883, 1431655765, 2863311531, ....
127094
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
385877
formula=(100+96+92+.........)-(98+94+90+.........)\nsum of arithmetic progression=Sn=(1/2)n[2a+(n-1)d]\n(1/2)*25{[200+24*2]-[196+24*2]}\n(1/2)*25*(248-244)\n(1/2)*25*4\n=50
13684
2+22+222+2222.... \n= 2 (1+11+111+1111...)\n= 2/9 (9+99+999+9999...)\n= 2/9 ( 10-1 + 100-1 + 1000-1 + ...)\n= sum of 2/9 (10^n -1)\n\nof course i worked backwards after looking at the answer, and others were correct as well. i just put it here.
18713
Looking at...\n\n>>>>>>>>> 0\n>>>>>>>>> 1\n>>>>>>>>> 2\n>>>>>> etc.\n>>>> 999999\n\n...in the ones or 10^0 column you repeat\n\n>>>> 0+1+2+3+4+5+6+7+8+9 = 45\n\n100,000 times.  That gives you 4,500,000.\n\nIn the tens or 10^1 column you repeat\n\n>>>> 10*0 + 10*1 + 10*2 + ... + 10*9 = 10*45\n\n10,000 times.  (i.e., group ten 0's in a row, then ten 1's in a row, ..., ten 9's in a row, and repeat that 10,000 times.)  That also gives you 4,500,000.\n\nLikewise with the 10^2, 10^3, 10^4, and 10^5 columns.\n\nSo far that totals 6 times 4,500,000 = 27,000,000.\n\nAll that remains is to add in the 1 from 1,000,000 to get the final total of 27,000,001.\n\nCheck: compare to previous reply. :-)\n\n(How do you get this thing to indent a line?)
596152
First, get rid of the brackets (multiply the terms inside the brackets by -1):\n\n7-2x^2 +2x -2 -2x -5 +2x^2\n\nCollect like terms:\n\n2x^2 -2x^2 +2x -2x +7-2-5 = 0\n\nIt's easy to see the sum is zero.
209883
Your equation would be:\n\n= ((1/p) - (1/q))/(1/pq)\n\n= ((q - p)/pq) * pq\n\n= q - p\n\nHope I am clear.
252985
Hopefully, I'm not doing your homework for you, and you are just curious about how to find this answer. Follow along on a piece of paper so it is easier to understand.\n\nIn a 12 hour period of time, reguardless of what time you start the count, only 1:00 through 6:00 can possibly be used in order to stay within the "6" limit. Let's see how many numbers will add to 6 in 1:00. \n\n1:05, 1:50, 1:14, 1:41, 1:23, 1:32. (6 Total)\n\nNow for the rest of the hours:\n\n2:04, 2:40, 2:13, 2:31, 2:22. (5 Total)\n3:03, 3:30, 3:12, 3:21. (4 Total)\n\nSee a pattern developing?\n\n4:02, 4:20, 4:11. (3 Total)\n5:01, 5:10. (2 Total)\n6:00. (1 Total)\n\nNow add the totals to get 21 times that the digits add up to 6!\nCongratulations.
665357
decimal 2+2=4\nbinary 2+2=10+10=100\noctal    2+2=4\nhexadecimal  2+2=4\n\n\nand if the base is 3 then  2+2=10
482189
Look at it like this:\n\n100 + 1 = 101\n99 + 2 = 101\n98 + 3 = 101\n97 + 4 = 101\n...\n\nSo, if you add all of the 101's up, you have 100 of them, or 100 * 101.  But you added 100 twice, 99 twice, 98 twice, ... 2 twice and 1 twice.  So, you divide by 2 to get the final answer:\n\n100 * 101 / 2 = 5050
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.