Challenging series problem for the intellectuals out there!?
410190
blah blahblah what the heck i'll guess . . . 1223494873984739847!!!!!!!!!!
482539
24\n\nthe series have u seen it
580109
ans is 70\n(5*2+1) (11*2+2) (24*2+3)\nrule is there r two series 5 11 24 51...\n7 16 34 70\n(7*2+2) (16*2+2) (34*2+2)
342433
1093 then 3280\n\nThis series is formed by adding 3 then 3x3 then 3x3x3 then 3x3x3x3 and so on
753727
Looks as though this is some kind of assignment. No? Maybe you ought to think it all through and provide your own answers. You can do it!!\n\nChow!!
200234
if this series goes upto infinity then just find the nth term which will be 2n-4.4 \nthen let E represent sigma notation\n∞\nE 2n-4.4 \nn=1\n\ni hope i was help
1043703
The ... implies an endless continuation of the series.  So the answer is \n\nINFINITY.
649544
Did you mean infinite series by the ellipsis? Then all these series obviously diverge to +infinity...
635766
Even the answer is not clear!
52450
There is no answer.\n\nWhat you are expressing is the sum of ((-1)^n+1 * 2) from 0 to infinity. This is a divergent series and therefor has no answer, or in mathematica terms, the answer DNE (does not exist).
650155
simple\nfirst\nwhen we apply ∑ we can write the series as\n(a_2 - a_1) + (a_3 - a_2) + (a_4 - a_3).... (a_n - a_n-1) (a_n+1 - a_n)\n\nwhen u add \nthe terms starting frm a_2(except a_1 & a_n+1) get cancelled with the correseponding terms and u get\na_n+1 + a_1
306730
well we know the 3 space must be even, in order to avoid the last space resulting in another 5\n\nhmmm... i'l work on it, provided you're working as well!\n\n..and #3 mustn't be 0, to avoid another 0\n..lol i'd have had it if i had been home...i made a program to do it, and then left before i finished/ran it lol\n\nbut that, is of course, the correct asnwer, 396*45=17820\n\nhaha thanx for the challenge!
978661
Have fun:\nhttp://www.math.nyu.edu/~griffith/teaching/06numanalysis/labs/bisection/
942547
Can you please check you haven't made a mistake somewhere? I have just calculated (with 99% certainty) that there are no solutions to those equations with those restrictions.\n\nEdit - OK, I assumed you made exactly one mistake. By removing each possible word and trying to solve it, I discovered that removing 'SIGH' was the only option that gave an answer.\nt 1\nh 8\na 3\nn 7\nk 6\ns 5\ng 9\ni 4\nv 2\nwhere SIGH = 26, not 25. Then THANKSGIVING = 65.\n\nThat was actually a cooler puzzle that the original, assuming you made one mistake :)
937394
Let the first term of the series be a.Therefore,the second term would be ar,r being the common ratio.and the fifth term is ar^4\nDividing ,the fifth term  ar^4 by the second term ar,we get.\nar^4/ar=(81/64)/3  or,r^3=27/64.Therefore r=3/4.\nPutting the valu of r in the secon term i.e. ar=3,we get a=3/r=3/(3/4)=4.\nTherefore,the series is 4,3......81/64.......\nit is apparent that the trms of the series would have the values increasing and it would not come to an end because it has attained the value of  more than an unity.\nThe sum of the series would be infinity.\nAs the sum of the series is infinity,the sum of firs 10 trms of sum of any number of trms for that matter,and the sum of the complete sries would be infinity as any number subtratd from infinity gives  result of infinity.
731608
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
858552
Have you tried these?  Surely you've been told that division by 0 isn't possible, so there's your answer for 7.  I'm not doing any others on Yahoo answers (too many at once), but if you really need help and are willing to do as much as possible for yourself I'll help you if you email h_chalker@yahoo.com.au, but nasty Maths teachers like me won't just hand you 7 answers on a plate!