find k and n if $(1+kx)^n=1-12x+60x^2......$

Question

If $(1 + kx)^n = 1 - 12x + 60x^2 - . . .$ what are the values of k and n ?

while the answer is supposedly k=-2 and n=6, I have gotten different solutions and would like someone to point out my errors

My steps:

$\binom n1(1)^{n-1}(kx)^1=-12$

which then becomes $(n)(k)=12$

and $\binom n2(1)^{n-2}(kx)^2=60x^2$

which becomes $(0.5)(n^2-n)(k)=60$

clearly when using substitution I should be getting $n=-9 $

I was wondering if the answer key was wrong or if there is truly some mistake that I am making?

[I would also like to have some pointers of how and where I have made the mistake]

Answer 1

You are on the right track, but made a few computational mistakes on the way. The firsts coefficients of $(1+kx)^n$ are $$(1+kx)^n=1+\binom n1 kx+\binom n2k^2x^2+\cdots$$ By identification we get the relations $$\binom n1k=-12 \quad \text{and}\;\;\binom n2k^2=60.$$ or by expanding the coefficients (using $\binom n2=\frac{n(n-1)}{2}$) $$nk=-12 \quad \text{and}\;\; n(n-1)k^2=120.$$ Since $n(n-1)k^2=n^2k^2-nk^2$, we get $nk^2=(-12)^2-120=24$. and we can deduce immediately that $k=24/(-12)=-2$, thus $n=6$.

Answer 2

Your second equation should be $$\frac{n(n-1)}2k^2=60.$$ Dividing by $nk=-12$ gives $$\frac{n-1}2k=-5$$ or $$nk-k=-10.$$ Things should be easy now.

Answer 3

Alternative solution in case you weren't familiar with binomial theorem.

$$(1+kx)^n = 1-12x+60x^2\cdots$$

Differentiating with respect to $x$

$$n(1+kx)^{n-1}k=-12+120x\cdots$$

Substituting $x=0$

$$n(1)k=-12\implies nk=-12$$

Differentiating again

$$n(n-1)(1+kx)^{n-2}k^2=120+\cdots$$

Substituting $x=0$

$$n(n-1)k^2=120$$