Binomial theprem and expansion

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    Binomial Expansion
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Discussion Overview

The discussion revolves around the binomial theorem and its expansion, specifically focusing on the last few terms of the expansion of (1+x)^n. Participants explore the formulation and implications of the binomial coefficients within this context.

Discussion Character

  • Exploratory, Homework-related

Main Points Raised

  • One participant requests clarification on the last few terms of the binomial expansion for (1+x)^n, indicating difficulty in finding a textbook answer.
  • Another participant notes the symmetry of binomial coefficients (nCr = nCn-r) and provides the last few terms of the expansion as + n(n-1) xn-2/2! + nxn-1/1! + xn.
  • A participant questions whether the binomial expansion can be used to prove the inequality (1+n)n ≥ 5/2nn- 1/2n n-1, seeking guidance on this proof.
  • There is a repeated request for assistance in writing out the last three terms of the expansion to aid in the proof of the mentioned inequality.

Areas of Agreement / Disagreement

The discussion remains unresolved, with participants expressing different needs for clarification and proof without reaching a consensus on the last terms or the proof's validity.

Contextual Notes

There are limitations in the clarity of the mathematical steps and assumptions regarding the inequality proof, which remain unresolved.

rohan03
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(1+x)^n=1+nx/1!+(n(n-1) x^2)/2!+⋯+ what are the last few terms of this ? I looked and tried but don't seem to get any textbook answer for this.
 
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hi rohan03! :smile:

(try using the X2 and X2 buttons just above the Reply box :wink:)

the binomial coefficients are symmetric (nCr = nCn-r),

so it ends … + n(n-1) xn-2/2! + nxn-1/1! + xn :wink:​
 
Thank you . Can I prove with the help of this :
(1+n)n ≥ 5/2nn- 1/2n n-1
 
and if yes - please guide on how
 
rohan03 said:
Thank you . Can I prove with the help of this :
(1+n)n ≥ 5/2nn- 1/2n n-1

just write out the last three terms …

what do you get? :smile:​
 
This was also posted under "homework" so I am closing this thread.
 

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