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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, 1 month ago

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Hi,

$$a_n = \dfrac{2 n^3 + 9 n^2 + n + 24}{ 6}$$ for n = 0,1,2 ... · 4 years, 1 month ago

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how did u get the answer? · 4 years, 1 month ago

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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} · 4 years, 1 month ago

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Yes, you are starting from n = 1. Thanks for the problem. · 4 years, 1 month ago

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i have a new one..........try this · 4 years, 1 month ago

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bj,bn · 8 months ago

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kh,l · 8 months ago

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vbhj · 8 months ago

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czfh · 8 months ago

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g · 8 months 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 · 4 years, 1 month ago

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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? · 4 years, 1 month ago

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you did the job rightly...... thanks for describing for him.... · 4 years, 1 month ago

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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... · 4 years, 1 month ago

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3rd order · 4 years, 1 month ago

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this is 3rd order and formula is (2n^3+3n^2-11n+30)/6 · 4 years, 1 month 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 · 4 years, 1 month ago

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the nth term is perhaps t=(2n^3+3n^2-11n+30)/6 · 4 years, 1 month ago

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its 4th order,still working on the formula · 4 years, 1 month ago

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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.... · 4 years, 1 month ago

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still working....!!??!!??....:) · 4 years, 1 month ago

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Lagrange Interpolation gives the lowest order polynomial that describes a group of points. · 4 years, 1 month ago

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I'm not so good at progressions. · 4 years, 1 month ago

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o.k. you are not jack of all trades..i.e. master of any special branch surely...have a try for it · 4 years, 1 month ago

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