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 Sayan Chaudhuri
8 years, 1 month ago

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12 votes

  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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Comments

Hi,

an=2n3+9n2+n+246a_n = \dfrac{2 n^3 + 9 n^2 + n + 24}{ 6} for n = 0,1,2 ...

gopinath no - 8 years, 1 month ago

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how did u get the answer?

swapnil mehta - 8 years, 1 month ago

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}

Sayan Chaudhuri - 8 years, 1 month ago

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

gopinath no - 8 years, 1 month ago

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@Gopinath No i have a new one..........try this

Sayan Chaudhuri - 8 years, 1 month ago

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 - 8 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: 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: 5,7,9,115,7,9,11 Again it is not of much meaning so find the differences between successive numbers in 5,7,9,115,7,9,11 which tantamounts to: 2,2,2,2,2,2,. 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 f(n)=an^3+bn^2+cn+d To find a,b,c,da,b,c,d just substitute for n=1,2,,3,4,5,6... Now u will know what to do?

Aditya Parson - 8 years, 1 month ago

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

Sayan Chaudhuri - 8 years, 1 month ago

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...

Sayan Chaudhuri - 8 years, 1 month ago

g

Galen Buhain - 4 years, 8 months ago

czfh

Galen Buhain - 4 years, 8 months ago

vbhj

Galen Buhain - 4 years, 8 months ago

kh,l

Galen Buhain - 4 years, 8 months ago

bj,bn

Galen Buhain - 4 years, 8 months ago

its 4th order,still working on the formula

Tan Li Xuan - 8 years, 1 month ago

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

Sayan Chaudhuri - 8 years, 1 month ago

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

Sebastian Garrido - 8 years, 1 month ago

I'm not so good at progressions.

Tan Li Xuan - 8 years, 1 month 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

Sayan Chaudhuri - 8 years, 1 month ago

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

Sayan Chaudhuri - 8 years, 1 month ago

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

saptami chattaraj - 8 years, 1 month ago

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 - 8 years, 1 month ago

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

bhuvnesh goyal - 8 years, 1 month ago

3rd order

Nadim Badi - 8 years, 1 month ago
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