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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)=an3+bn2+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
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
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2^{34}
a_{i-1}
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Comments
Hi,
an=62n3+9n2+n+24 for n = 0,1,2 ...
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how did u get the answer?
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}
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Yes, you are starting from n = 1. Thanks for the problem.
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this
i have a new one..........tryNice 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)=an3+bn2+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....
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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its 4th order,still working on the formula
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still working....!!??!!??....:)
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Lagrange Interpolation gives the lowest order polynomial that describes a group of points.
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 foractually it is 3rd order...very few of you showed interest in the problem.... better try tomorrow's problem....and find nth order....
the nth term is perhaps t=(2n^3+3n^2-11n+30)/6
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
this is 3rd order and formula is (2n^3+3n^2-11n+30)/6
3rd order