coefficients of $x^k$ and $x^{k+1}$

Question

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

How should I solve this question? Thanks in advance.

Answer 1

Note that the coefficient of $x^k$ in your polynomial is $\binom{19}{k}3^k2^{19-k}$ due to the binomial theorem.

You want that: $$\binom{19}{k}3^k2^{19-k} = \binom{19}{k+1}3^{k+1}2^{18-k}$$ We can expand the binomial coefficients to get that: $$\frac{19!}{k!(19-k)!} 3^k2^{19-k} = \frac{19!}{(k+1)k!(18-k)!}3^{k+1}2^{18-k}$$ Now, we try to cancel as much stuff as we can. $$\frac{2}{(19-k)!} = \frac{3}{(k+1)(18-k)!}\implies \frac{2}{19-k} = \frac{3}{k+1}\implies 2k+2 = 57-3k\implies k = 11$$