find n, if the coefficient of $x^2$ in $(3 + 2x)^n$ is 20412.

Question

Find $n$, if the coefficient of $x^2$ in $(3+2x)^n$ is 20412

I solved it using binomial theorem and got $n(n-1)3^{n-2} = 10206$

The term like $n(n-1)3^{n-2}$, I am not sure how to solve it.

Answer 1

The coefficient should be $2n(n-1)3^{n-2}$.

To obtain $n$ use prime factorisation: $\;20412=2^2\cdot3^6\cdot 7$. Maybe $n$ or $n-1$ is divisible by $3$, so the equality $$2n(n-1)3^{n-2}=2^2\cdot3^6\cdot 7$$ implies $\;n\le 8$.

By Euclid's lemma, it also implies $7$ divides $n$ or $n-1$. Combining with $n\le 8$, we have that $n=7$ or $n=8$. However $n=8$ is impossible, as the expression is divisible only by $2^2$. Thus there is only one solution: $\;n=\color{red}7$.

Answer 2

The question asks to find the $n$ so that $$ \binom{n}{2}3^{n-2}(2x)^2=20412x^2\tag{1} $$ That is $$ 2n(n-1)3^{n-2}=20412\tag{2} $$ Plugging in values of $n$ leads to a solution pretty quickly.

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We can also try to apply the Lambert W Function. Note that $n(n-1)\sim\left(n-\frac12\right)^2$. Therefore, multiplying $(2)$ by $\frac{3^{3/2}}2$, taking the square root, and multiplying by $\frac{\log(3)}2$, we want to solve $$ \left(n-\frac12\right)\frac{\log(3)}2\,e^{\left(n-\frac12\right)\frac{\log(3)}2}\approx\frac{\log(3)}2\sqrt{20412\cdot\frac{3^{3/2}}2}\tag{3} $$ That is, $$ \left(n-\frac12\right)\frac{\log(3)}2 \approx\operatorname{W}\left(\frac{\log(3)}2\sqrt{20412\cdot\frac{3^{3/2}}2}\right)\tag{4} $$ Computing $$ \begin{align} n &\approx\frac12+\frac2{\log(3)}\operatorname{W}\left(\frac{\log(3)}2\sqrt{20412\cdot\frac{3^{3/2}}2}\right)\\ &=6.9957802\tag{5} \end{align} $$ Checking $n=7$ in $(2)$ shows that that is the solution.

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Lacking an appropriate CAS, an algorithm is given in this answer to compute Lambert W.