See,
if you calculate the difference between the numbers(6 as displayed) it comes to \(:
2,7,14,23,34\).
Since we cannot see any meaning in the above derived sequence we find the difference between the successive numbers of the same which comes to:
\(5,7,9,11\)
Again it is not of much meaning so find the differences between successive numbers in \(5,7,9,11\) which tantamounts to:
\(2,2,2,\).
Hence we can conclude that the above sequence in the question is an equation of degree 3.
So we can now write in the form:
\( f(n)=an^3+bn^2+cn+d\)
To find \(a,b,c,d\) just substitute for n=1,2,,3,4,5,6...
Now u will know what to do?

here see Muhammad, i am not as expert in math as calvin,the problem master, is ....so when i come across some math problem that can make oneself think in a new process or supplies an additional data, i want to post them to spread among my net-friends ....so i could not meet your expectation to provide "proper step-wise solution" and i think you ,as being in university, could atleast get at the heart of the logic under this problem......enjoy the problems posted by me daily....thanking you...

Easy Math Editor

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boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

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## Comments

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TopNewestHi,

\(a_n = \dfrac{2 n^3 + 9 n^2 + n + 24}{ 6}\) for n = 0,1,2 ...

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how did u get the answer?

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yes you are right....according to my calculation....::::::another possible sequence can be identified as :: a_n = \frac {2

n^{3}+3n^{2}-11n+30}{6}Log in to reply

Yes, you are starting from n = 1. Thanks for the problem.

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this

i have a new one..........tryLog in to reply

bj,bn

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kh,l

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vbhj

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czfh

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g

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Nice problem but at least one should provide us the proper step-wise solution or at least logic behind the answer. I hope, you'd! Regards

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See, if you calculate the difference between the numbers(6 as displayed) it comes to \(: 2,7,14,23,34\). Since we cannot see any meaning in the above derived sequence we find the difference between the successive numbers of the same which comes to: \(5,7,9,11\) Again it is not of much meaning so find the differences between successive numbers in \(5,7,9,11\) which tantamounts to: \(2,2,2,\). Hence we can conclude that the above sequence in the question is an equation of degree 3. So we can now write in the form: \( f(n)=an^3+bn^2+cn+d\) To find \(a,b,c,d\) just substitute for n=1,2,,3,4,5,6... Now u will know what to do?

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you did the job rightly...... thanks for describing for him....

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here see Muhammad, i am not as expert in math as calvin,the problem master, is ....so when i come across some math problem that can make oneself think in a new process or supplies an additional data, i want to post them to spread among my net-friends ....so i could not meet your expectation to provide "proper step-wise solution" and i think you ,as being in university, could atleast get at the heart of the logic under this problem......enjoy the problems posted by me daily....thanking you...

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3rd order

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this is 3rd order and formula is (2n^3+3n^2-11n+30)/6

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it's a series where the 3rd differnce series is ap of c.d 4 so the formula for the n'th term is definitely {2n^3+3n^2-11n+30}/6

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the nth term is perhaps t=(2n^3+3n^2-11n+30)/6

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its 4th order,still working on the formula

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actually it is 3rd order...very few of you showed interest in the problem.... better try tomorrow's problem....and find nth order....

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still working....!!??!!??....:)

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Lagrange Interpolation gives the lowest order polynomial that describes a group of points.

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I'm not so good at progressions.

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it

o.k. you are not jack of all trades..i.e. master of any special branch surely...have a try forLog in to reply