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.
+680 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 
+9673 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.
