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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
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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 Jun 13, 2020
#2
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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