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dpflan
 2 years ago
Best ResponseYou've already chosen the best response.1Ah, this is a nice series. \[[5,8,17,44]\]

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.2find the difference .. note the difference in the difference is a multiple of 3

Yahoo!
 2 years ago
Best ResponseYou've already chosen the best response.0Yes the first order difference is in Gp

dpflan
 2 years ago
Best ResponseYou've already chosen the best response.1and on...so do as @experimentX suggests, first examine any sort of relationship that could exist, you see that each termer is greater than the next

dpflan
 2 years ago
Best ResponseYou've already chosen the best response.1I suggest that you show us the values you have for the difference between the numbers in the series

dpflan
 2 years ago
Best ResponseYou've already chosen the best response.1Once, you have those differences, see if there is a pattern or relationship among them

zepp
 2 years ago
Best ResponseYou've already chosen the best response.0Looks like 3 to some power :D

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.2tn+1 = tn + 3^(n1)

dpflan
 2 years ago
Best ResponseYou've already chosen the best response.1term1 = 5 term2 = 8 term2  term1 = 3 so term2 = term1 + 3 term3 = 17 term3  term2 = 9 so term3 = term2 + 9 term4 = 44 term5  term4 = 27 so term4 = term3 + 27 term5 = ...

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.2don't know closed form at the moment!!

dpflan
 2 years ago
Best ResponseYou've already chosen the best response.1So @Yahoo! How are you doing with this?

Yahoo!
 2 years ago
Best ResponseYou've already chosen the best response.0Ar^n + b1n^k1 + b2n^k2............bk

dpflan
 2 years ago
Best ResponseYou've already chosen the best response.1Hmm, can you explain the terms please?

Yahoo!
 2 years ago
Best ResponseYou've already chosen the best response.0it is the general formula lol1

sami21
 2 years ago
Best ResponseYou've already chosen the best response.0ok look at the sequence first term is (3^1+7)/2=5 second is (3^2+7)/2=8 third is (3^3+8)/2=17 can you now tell me general formula ?

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.2i guess we should use summation rule dw:1343147827410:dw

mahmit2012
 2 years ago
Best ResponseYou've already chosen the best response.1dw:1343147906405:dw

mahmit2012
 2 years ago
Best ResponseYou've already chosen the best response.1dw:1343147929358:dw

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.2not sure that works!

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.2dw:1343147968035:dw something like that!! just try to manipulate and fit in it ...

mahmit2012
 2 years ago
Best ResponseYou've already chosen the best response.1dw:1343147962155:dw

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.2dw:1343148038487:dw

mahmit2012
 2 years ago
Best ResponseYou've already chosen the best response.1dw:1343148023475:dw

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.0i like to get the first term to 1 alot of times 5 + 8 + 17 + 44 4 4 4 4  1 + 4 + 13 + 40 just helps me with the inbetween addings or if we up it by +1 5 + 8 + 17 + 44 +1 +1 +1 +1  6 + 9 + 18 + 45

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.0factor out a 3 3( 2 + 3 + 6 + 15 + ...) all ways to se it differently

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.2dw:1343148174027:dw

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.2i hope it works for other terms too .. let's see for sum

Yahoo!
 2 years ago
Best ResponseYou've already chosen the best response.0@experimentX i didnt understand!!

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.2dw:1343148311125:dw

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.2that's just some manipulation .. typing in latex really takes lot of time

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.2\[ t_n = t_{n1} + 3^{n1}\] this is the general rule \[ t_{n} = t_{n2} + 3^{n 2} + 3^{n1} \] continue this way and you will get \[ t_{n} = t_1 + 3 + 3^3 + 3^4+ .. + 3^{n 2} + 3^{n1} = 5 + 3{ 3^{n1}  1 \over 31}\] which is requited formula

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.2for summation \[ \sum_{n=1}^k {5 + {3 \over 2} (3^{n  1 }  1)} =\sum_{n=1}^k{7/2} + 3^n = k {7 \over 2}+ 3{3^k 1 \over 31} = k {7 \over 2}+ {3 \over 2}(3^k 1)\] @Yahoo! do you understand??

mahmit2012
 2 years ago
Best ResponseYou've already chosen the best response.1dw:1343149457407:dw

mahmit2012
 2 years ago
Best ResponseYou've already chosen the best response.1@experimentX I guess you missed 1/2 .

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.2thanks @mahmit2012 for correction ... you are sharp!!ii

Yahoo!
 2 years ago
Best ResponseYou've already chosen the best response.0@mahmit2012 @experimentX i learnt this by using this Ar^n + b1n^k1 + b2n^k2............bk

mahmit2012
 2 years ago
Best ResponseYou've already chosen the best response.1You're welcome , and good tutoring you have had. @experimentX

Yahoo!
 2 years ago
Best ResponseYou've already chosen the best response.0Ar^n + b1n^k1 + b2n^k2............bk

Yahoo!
 2 years ago
Best ResponseYou've already chosen the best response.0can u ssolve this by using Ar^n + b1n^k1 + b2n^k2............bk

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.2you are a good tutor!! most of me is autodidact!!
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