Binomial Expansion $(1+kx)^4 (1+x)^n$

Question

Question

In the expansion of $(1+kx)^4 (1+x)^n$, the coefficient of $x$ is $13$ and the coefficient of $x^2$ is $74$. Find the possible values of $k$ and $n$.

My solution

$$ \begin{split} (1+kx)^4 &= 1+4kx+6k^2 x^2+4k^3 x^3+k^4 x^4\\ (1+x)^n &= 1+ \binom{n}{1}x+\binom{n}{2} x^2... \end{split} $$

So $\binom{n}{1}x + 4kx=13x$ and $\binom{n}{2}x^{2}+4k\binom{n}{1}x^{2}+6k^{2}x^{2}=74x^{2}$.

That is, $$\binom{n}{1}+4k=13 \quad (*)$$ and $$\binom{n}{2}+4k\binom{n}{1}+6k^{2}=74 \quad (**)$$

Since $\binom{n}{1}=n$, we have $n+4k=13$.

From trial and improvement I worked out that $n = 5$ and $k = 2$ satisfies both equations (*) and (**) above.

How do I find the other set of solutions?

Answer 1

The second equation gives $$n(n-1) + 8nk + 12k^2 = 148$$ (by multiplying the equation by $2$ and noting $\binom n2 = \frac{n(n-1)}2.$

We have $n+4k=13$. Write $n = 13-4k$, and substitute this into the above equation, to obtain $$(13-4k)(12-4k) + 8(13-4k)k + 12k^2 = 148,$$ which is a quadratic in $k$. You can now solve for $k$, and then use $n=13-4k$ to find $n$.