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Hello Solve this interesting problem

From today I will discuss one interesting math problem in this forum daily...keep in touch...try to solve them.

Note by Raja Metronetizen
4 years, 3 months ago

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Hi,

\(a_n = \dfrac{2 n^3 + 9 n^2 + n + 24}{ 6}\) for n = 0,1,2 ... Gopinath No · 4 years, 3 months ago

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@Gopinath No how did u get the answer? Swapnil Mehta · 4 years, 3 months ago

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@Gopinath No yes you are right....according to my calculation....::::::another possible sequence can be identified as :: a_n = \frac {2n^{3}+3n^{2}-11n+30}{6} Raja Metronetizen · 4 years, 3 months ago

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@Raja Metronetizen Yes, you are starting from n = 1. Thanks for the problem. Gopinath No · 4 years, 3 months ago

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@Gopinath No i have a new one..........try this Raja Metronetizen · 4 years, 3 months ago

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bj,bn Galen Buhain · 10 months, 3 weeks ago

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kh,l Galen Buhain · 10 months, 3 weeks ago

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vbhj Galen Buhain · 10 months, 3 weeks ago

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czfh Galen Buhain · 10 months, 3 weeks ago

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g Galen Buhain · 10 months, 3 weeks ago

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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 Muhammad Abdullah · 4 years, 3 months ago

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@Muhammad Abdullah 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? Aditya Parson · 4 years, 3 months ago

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@Aditya Parson you did the job rightly...... thanks for describing for him.... Raja Metronetizen · 4 years, 3 months ago

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@Muhammad Abdullah 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... Raja Metronetizen · 4 years, 3 months ago

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3rd order Nadim Badi · 4 years, 3 months ago

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this is 3rd order and formula is (2n^3+3n^2-11n+30)/6 Bhuvnesh Goyal · 4 years, 3 months ago

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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 Priyajit Ghosh · 4 years, 3 months ago

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the nth term is perhaps t=(2n^3+3n^2-11n+30)/6 Saptami Chattaraj · 4 years, 3 months ago

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its 4th order,still working on the formula Tan Li Xuan · 4 years, 3 months ago

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@Tan Li Xuan actually it is 3rd order...very few of you showed interest in the problem.... better try tomorrow's problem....and find nth order.... Raja Metronetizen · 4 years, 3 months ago

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@Tan Li Xuan still working....!!??!!??....:) Raja Metronetizen · 4 years, 3 months ago

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@Raja Metronetizen Lagrange Interpolation gives the lowest order polynomial that describes a group of points. Sebastian Garrido · 4 years, 3 months ago

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@Raja Metronetizen I'm not so good at progressions. Tan Li Xuan · 4 years, 3 months ago

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@Tan Li Xuan o.k. you are not jack of all trades..i.e. master of any special branch surely...have a try for it Raja Metronetizen · 4 years, 3 months ago

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