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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, 6 months ago

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12 votes

  Easy Math Editor

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**bold** or __bold__ bold

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[example link](https://brilliant.org)example link
> This is a quote
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    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 \( 2 \times 3 \)
2^{34} \( 2^{34} \)
a_{i-1} \( a_{i-1} \)
\frac{2}{3} \( \frac{2}{3} \)
\sqrt{2} \( \sqrt{2} \)
\sum_{i=1}^3 \( \sum_{i=1}^3 \)
\sin \theta \( \sin \theta \)
\boxed{123} \( \boxed{123} \)

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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, 6 months ago

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how did u get the answer?

Swapnil Mehta - 4 years, 6 months 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}

Raja Metronetizen - 4 years, 6 months ago

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

Gopinath No - 4 years, 6 months ago

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

Raja Metronetizen - 4 years, 6 months ago

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bj,bn

Galen Buhain - 1 year, 1 month ago

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kh,l

Galen Buhain - 1 year, 1 month ago

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vbhj

Galen Buhain - 1 year, 1 month ago

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czfh

Galen Buhain - 1 year, 1 month ago

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g

Galen Buhain - 1 year, 1 month 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, 6 months 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?

Aditya Parson - 4 years, 6 months ago

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

Raja Metronetizen - 4 years, 6 months 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...

Raja Metronetizen - 4 years, 6 months ago

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3rd order

Nadim Badi - 4 years, 6 months ago

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

Bhuvnesh Goyal - 4 years, 6 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, 6 months ago

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

Saptami Chattaraj - 4 years, 6 months ago

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its 4th order,still working on the formula

Tan Li Xuan - 4 years, 6 months 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....

Raja Metronetizen - 4 years, 6 months ago

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

Raja Metronetizen - 4 years, 6 months ago

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

Sebastian Garrido - 4 years, 6 months ago

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I'm not so good at progressions.

Tan Li Xuan - 4 years, 6 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, 6 months ago

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