binomial theorem relationship between coefficients (2 Viewers)

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 (2 Viewers)

cos16mh

New Member

Drongoski

Well-Known Member

cos16mh

New Member

Drongoski

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 2)