If the coefficients of $x^{k}$ and $x^{k+1}$ in the expansion of $(2+3x)^{19}$ are equal, how to find $k$?

Question

If the coefficients of $x^{k}$ and $x^{k+1}$ in the expansion of $(2+3x)^{19}$ are equal, find $k$.

Answer 1

The coefficient of $a^{19-k}b^k$ in the expansion of $(a+b)^{19}$ is $\binom{19}{k}$

This implies for your specific problem by setting $a=2$ and $b=3x$ then that the coefficient of $x^k$ in the expansion of $(2+3x)^{19}$ is

$2^{19-k}3^k\binom{19}{k}$

Knowing this, and that the coefficients of $x^k$ and $x^{k+1}$ are equal, we have the following equation:

$2^{19-k-1}3^{k+1}\binom{19}{k+1}=2^{19-k}3^k\binom{19}{k}$

Algebraically manipulating this will lead you to an answer.

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One more step: Technically, to continue properly, we should assume that $0\leq k\leq 18$ to avoid either side equaling zero so we can multiply and divide by things without fear of division by zero errors.

Dividing both sides by $2^{19-k}3^k$ and replacing binomial coefficients with factorials, we get $\frac{3}{2}\frac{19!}{(k+1)!(19-k-1)!}=\frac{19!}{k!(19-k)!}$

Continue by cancelling what you can by recognizing some key defining properties about factorials, most importantly that $(k+1)!=(k+1)\cdot k!$ to try to find what $k$ must be equal to.