binomial theorem relationship between coefficients (1 Viewer)

cos16mh · Feb 21, 2018

Hi I'm stuck on how to begin this question Assistance would be appreciated

The coefficients of x^4 and x^5 in the expansion of (3 - x)^n are equal in magnitude but opposite in sign. Find the value of n.

Drongoski · Feb 21, 2018

cos16mh said:
Hi I'm stuck on how to begin this question Assistance would be appreciated

The coefficients of x^4 and x^5 in the expansion of (3 - x)^n are equal in magnitude but opposite in sign. Find the value of n.

$\therefore $ coeff of $ x^4 = - $(coeff of $ x^5 ) \\ \\ \therefore \binom n 4 3^{n-4} (-1)^4 = - (\binom n 5 3^{n-5} (-1)^5) \\ \\ \therefore \binom n 4 3^{n-4}= \binom n 5 3^{n-5 }\\ \\ \therefore 3\times \binom n 4 = \binom n 5 \\ \\ \therefore \frac {3 \times n!}{4! \times (n-4) \times (n-5)!} = \frac {n!}{5 \times 4! \times (n-5)!}$

$\therefore \frac {3}{n-4} = \frac {1}{5}\\ \\ \therefore n = 19$

cos16mh · Feb 22, 2018

Thanks could you talk me through from the third last step to the second last step please

Drongoski · Feb 22, 2018

cos16mh said:
Thanks. Could you talk me through from the third last step to the second last step please

No problem. I'll walk you thru those steps.

$\binom n 4 = \frac {n!}{4! (n-4)!} \\ \\ (n-4)! = 1 \times 2 \times 3 \times . . . \times (n-5) \times (n-4) \\ \\ = (n-4) \times \left ( 1 \times 2 \times 3 \times . . . \times (n-5) \right ) = (n-4) \times (n-5)!$

5! = 5 x 4!

You cancel all the common factors on both sides of the equation to arrive at the 2nd last line.

binomial theorem relationship between coefficients (1 Viewer)

cos16mh

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Drongoski

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cos16mh

New Member

Drongoski

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