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Find the n th term and sum of n terms of the series 5 + 8 + 17 + 44 ..........
 one year ago
 one year ago
Find the n th term and sum of n terms of the series 5 + 8 + 17 + 44 ..........
 one year ago
 one year ago

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dpflanBest ResponseYou've already chosen the best response.1
Ah, this is a nice series. \[[5,8,17,44]\]
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
find the difference .. note the difference in the difference is a multiple of 3
 one year ago

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

dpflanBest ResponseYou've already chosen the best response.1
and 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
 one year ago

dpflanBest ResponseYou've already chosen the best response.1
I suggest that you show us the values you have for the difference between the numbers in the series
 one year ago

dpflanBest ResponseYou've already chosen the best response.1
Once, you have those differences, see if there is a pattern or relationship among them
 one year ago

zeppBest ResponseYou've already chosen the best response.0
Looks like 3 to some power :D
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
tn+1 = tn + 3^(n1)
 one year ago

dpflanBest ResponseYou've already chosen the best response.1
term1 = 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 = ...
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
don't know closed form at the moment!!
 one year ago

dpflanBest ResponseYou've already chosen the best response.1
So @Yahoo! How are you doing with this?
 one year ago

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

dpflanBest ResponseYou've already chosen the best response.1
Hmm, can you explain the terms please?
 one year ago

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

sami21Best ResponseYou've already chosen the best response.0
ok 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 ?
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
i guess we should use summation rule dw:1343147827410:dw
 one year ago

mahmit2012Best ResponseYou've already chosen the best response.1
dw:1343147906405:dw
 one year ago

mahmit2012Best ResponseYou've already chosen the best response.1
dw:1343147929358:dw
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
not sure that works!
 one year ago

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

mahmit2012Best ResponseYou've already chosen the best response.1
dw:1343147962155:dw
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
dw:1343148038487:dw
 one year ago

mahmit2012Best ResponseYou've already chosen the best response.1
dw:1343148023475:dw
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
i 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
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
factor out a 3 3( 2 + 3 + 6 + 15 + ...) all ways to se it differently
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
dw:1343148174027:dw
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
i hope it works for other terms too .. let's see for sum
 one year ago

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

experimentXBest ResponseYou've already chosen the best response.2
dw:1343148311125:dw
 one year ago

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

experimentXBest 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
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
for 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??
 one year ago

mahmit2012Best ResponseYou've already chosen the best response.1
dw:1343149457407:dw
 one year ago

mahmit2012Best ResponseYou've already chosen the best response.1
@experimentX I guess you missed 1/2 .
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
thanks @mahmit2012 for correction ... you are sharp!!ii
 one year ago

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

mahmit2012Best ResponseYou've already chosen the best response.1
You're welcome , and good tutoring you have had. @experimentX
 one year ago

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

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

experimentXBest ResponseYou've already chosen the best response.2
you are a good tutor!! most of me is autodidact!!
 one year ago
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