Binomial Theorem

**kangta27** · January 5th 2012, 11:00 AM

In the expansion of (x+y)^n the coefficient of x^4y^(n-4) is 3,876 and the coefficient of x^5y^(n-5) is 11,628. Find the coefficient of x^5y^(n-4) in the expansion of (x+y)^(n+1). What is the value of n?

**Ridley** · January 5th 2012, 03:20 PM

Each coefficient can be calculated as follows: $a_i=\frac{n!}{i!(n-i)!}$

$a_4=\frac{n!}{4!(n-4)!}=3876 \Leftrightarrow n! = 3876 \cdot 4! \cdot (n-4)!$ (1)
$a_5=\frac{n!}{5!(n-5)!}=11628 \Leftrightarrow n! = 11628 \cdot 5! \cdot (n-5)!$ (2)

Combine (1) and (2):
$3876 \cdot 4! \cdot (n-4)! = 11628 \cdot 5! \cdot (n-5)! \Leftrightarrow n-4 = \frac{11628 \cdot 5!}{3876 \cdot 4!}$

**kangta27** · January 5th 2012, 03:39 PM

Thank you so much Ridley! My only question is, what happened to the (n-4)! and (n-5)! when you combined (1) and (2)?

**Ridley** · January 5th 2012, 03:44 PM

$(n-4)!= (n-4)\cdot(n-5)\cdot(n-6)\cdot...\cdot1 = (n-4) \cdot (n-5)!$

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