Solve Binomial Expansion Homework: (a) (1-x6)4, (b), (c) |x|<1

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The discussion revolves around solving binomial expansion problems, specifically expanding (1-x^6)^4 and finding coefficients in related expressions. The correct expansion for part (a) is identified as 1 - 4x^6 + 6x^12 - 4x^18 + x^24. For part (b), the coefficient of x^r in the expansion of (1-x)^-4 is given by the formula (r+1)(r+2)(r+3)/6. The key challenge lies in part (c), where the coefficient of x^8 in the expansion of ((1-x^6)/(1-x))^4 is determined to be 125, achieved by combining terms appropriately. The conversation highlights the importance of understanding how to manipulate the expansions to find specific coefficients.
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Homework Statement



(a) Expand (1-x6)4
(b) Find the coefficient of xr, where r is a non-negative integer, in the expansion of (1-x)-4 for |x|<1.
(c) Using (a) and (b), or otherwise, find the coefficient of x^8 in the expansion of ((1-x6)/(1-x))4 for |x|<1.

(Answers:
(a) 1-4x6+6x12-4x18+x24
(b) (r+1)(r+2)(r+3)/6
(c) 125)

Homework Equations



Binomial Expansion: n(n-1)...(n-r+1)*xr/r!

The Attempt at a Solution



I can solve parts (a) and (b) but not part (c).

I tried (-4)(2+3)(2+2)(2+1)/6 = -40 but not correct.

Can anyone tell me how to solve it?

Thank you very much!
 
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Hi chrisyuen! :smile:
chrisyuen said:
(c) Using (a) and (b), or otherwise, find the coefficient of x^8 in the expansion of ((1-x6)/(1-x))4 for |x|<1.

(Answers:
(a) 1-4x6+6x12-4x18+x24
(b) (r+1)(r+2)(r+3)/6
(c) 125)

I tried (-4)(2+3)(2+2)(2+1)/6 = -40 but not correct.

Yes, that's x6 times x2

now what about 1 times x8? :wink:
 
tiny-tim said:
Hi chrisyuen! :smile:


Yes, that's x6 times x2

now what about 1 times x8? :wink:

Yes, I got it!

Thank you very much!
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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