Find the Coefficient of $x^{14}$ in $(3-2x^2)^{10}$

Question

Find the Coefficient of $x^{14}$ in $(3-2x^2)^{10}$

I plugged this into the Binomial Theorem but the result was a negative number nor was i confident it was correct.
I'm hopelessly confused, pls help.

Answer 1

A really useful formula to take from the Binomial Theorem is that the $k+1$-th term (sometimes referred to as the general term) from the expansion of $(x+y)^{n}$ is given by $$T_{k+1}=\begin{pmatrix}n\\k\end{pmatrix}x^{n-k}y^{k}$$

For your problem, we have the expansion of $\left(3-2x^{2}\right)^{10}$ and we want to find the coefficient of $x^{14}$. The general term expression is thus: $$T_{k+1}=\begin{pmatrix}10\\k\end{pmatrix}\left(3\right)^{10-k}\left(-2x^{2}\right)^{k}$$

From here, you can identify that you get $x^{14}$ when $k=7$ since you'll have $\left(-2x^{2}\right)^{7}=\left(-2\right)^{7}x^{14}$.

Plugging $k=7$ into the general term expression gives us: $$\begin{align*} T_{8} &=\begin{pmatrix}10\\7\end{pmatrix}\left(3\right)^{10-7}\left(-2x^{2}\right)^{7}\\ &=\frac{10!}{(10-7)!\cdot 7!}3^{3}\left(-2^{7}\right)x^{14} \end{align*}$$

From this, we can evaluate the coefficient of $x^{14}$ to be: $$\begin{align*} \frac{10!}{(10-7)!\cdot 7!}3^{3}\left(-2^{7}\right) &=120\times27\times(-128)\\ &=-414720 \end{align*}$$

Answer 2

$$(3-2x^2)^{10} = {10 \choose 0}(3)^{10}-{10 \choose 1}(3)^9(2x^2) + {10 \choose 2}(3)^8(2x^2)^2-{10 \choose 3}(3)^7(2x^2)^3+...{10 \choose 10}(2x^2)^{10}$$ $x^{14}$ would mean $(x^2)^7$, so it’s the $8$th term.

It becomes $$-{10 \choose 7}(3)^3(2x^2)^7$$ $$\implies -414720x^{14}$$ So, the coefficient of $x^{14}$ is $-414720$.

Answer 3

Pedestrian:

$(-2x^2+3)^{10}=$

$\binom{10}{0}(-2x^2)^{10}3^0 +\binom{10}{1}(-2x^2)^93^1+$

$\binom{10}{2}(-2x^2)^83^2 +$

$\binom{10}{3}(-2x^2)^73^3.......$