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

Mathematics
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Ah, this is a nice series. \[[5,8,17,44]\]
find the difference .. note the difference in the difference is a multiple of 3
Yes the first order difference is in Gp

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Other answers:

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
I suggest that you show us the values you have for the difference between the numbers in the series
Once, you have those differences, see if there is a pattern or relationship among them
Looks like 3 to some power :D
tn+1 = tn + 3^(n-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 = ...
don't know closed form at the moment!!
So @Yahoo! How are you doing with this?
Ar^n + b1n^k-1 + b2n^k-2............bk
Hmm, can you explain the terms please?
it is the general formula lol1
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 ?
i guess we should use summation rule |dw:1343147827410:dw|
|dw:1343147906405:dw|
|dw:1343147929358:dw|
not sure that works!
|dw:1343147968035:dw| something like that!! just try to manipulate and fit in it ...
|dw:1343147962155:dw|
|dw:1343148038487:dw|
|dw:1343148023475:dw|
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
factor out a 3 3( 2 + 3 + 6 + 15 + ...) all ways to se it differently
|dw:1343148174027:dw|
i hope it works for other terms too .. let's see for sum
@experimentX i didnt understand!!
|dw:1343148311125:dw|
that's just some manipulation .. typing in latex really takes lot of time
\[ t_n = t_{n-1} + 3^{n-1}\] this is the general rule \[ t_{n} = t_{n-2} + 3^{n -2} + 3^{n-1} \] continue this way and you will get \[ t_{n} = t_1 + 3 + 3^3 + 3^4+ .. + 3^{n -2} + 3^{n-1} = 5 + 3{ 3^{n-1} - 1 \over 3-1}\] which is requited formula
for summation \[ \sum_{n=1}^k {5 + {3 \over 2} (3^{n - 1 } - 1)} =\sum_{n=1}^k{7/2} + 3^n = k {7 \over 2}+ 3{3^k -1 \over 3-1} = k {7 \over 2}+ {3 \over 2}(3^k -1)\] @Yahoo! do you understand??
|dw:1343149457407:dw|
@experimentX I guess you missed 1/2 .
thanks @mahmit2012 for correction ... you are sharp!!ii
@mahmit2012 @experimentX i learnt this by using this Ar^n + b1n^k-1 + b2n^k-2............bk
formula!1
You're welcome , and good tutoring you have had. @experimentX
Ar^n + b1n^k-1 + b2n^k-2............bk
can u ssolve this by using Ar^n + b1n^k-1 + b2n^k-2............bk
you are a good tutor!! most of me is autodidact!!

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