+0  
 
0
81
2
avatar

Find the value of n if the coefficients of x^3 and x^4 are equal in the expansion of (2x+1)^n

 

 

Any help or hints are greatly appreciated <3.

 Jun 12, 2020
 #1
avatar+549 
+1

Find the value of n if the coefficients of \(x^3\) and \(x^4\)are equal in the expansion of \((2x+1)^n\).

I rewrote bc i have a pet peeve where i don't like it when the LaTeX is wrong

it's nothing personal laugh

 Jun 13, 2020
 #2
avatar+8341 
0

We use binomial theorem.

 

\((2x + 1)^n = \displaystyle\sum_{k = 0}^n \binom nk (2x)^k\)

 

Coefficient of \(x^3\) is \(\displaystyle \binom n3 (2)^3 = \dfrac{4n(n - 1)(n - 2)}{3}\) for positive integer n.

 

Coefficient of \(x^4\) is \(\displaystyle \binom n4 (2)^4 = \dfrac{2n(n - 1)(n - 2)(n - 3)}{3}\) for positive integer n.

 

Now, the coefficient of \(x^3\) is equal to that of \(x^4 \).

 

\(\dfrac{4n(n - 1)(n - 2)}{3} = \dfrac{2n(n - 1)(n - 2)(n - 3)}{3}\\ 2n(n - 1)(n - 2) = n(n - 1)(n - 2)(n - 3)\\ n(n - 1)(n - 2)(n - 5) = 0\\ n = 0 \text{ (rej.) or }n = 1 \text{ (rej.) or } n = 2 \text{ (rej.) or } n = 5\)

 

Therefore n = 5.

 Jun 13, 2020

16 Online Users