Find coefficient of $X^{12}$

Question

I need to find coefficient of $X^{12}$ in $({1-2X})^{19}$.

What is the formula to solve it?I only know about

$$\frac{1}{a-X}=\frac{1}{a}\sum_{r=0}^\infty \frac{X^r}{a^r}$$ $$\frac{1}{(1-aX)^{n}}=\sum_{r=0}^\infty {X^r}{a^r} C(n-1+r,r)$$

Answer 1

You have to use the binomial theorem:

$(a+b)^n=\sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^k$

In this case you have $a=1$, $b=-2X$ and $n=19$

Therefore the term which will have $X^{12}$, is the term in which $k=12$, therefore, for this $k$, we have that

$[X^{12}]=\binom{19}{12} 1^{19-12}(-2)^{12}=\binom{19}{12} (-2)^{12}$

Here $[X^{12}]$ is the coefficient of $X^{12}$ in the expansion.

Answer 2

You don't have to use the binomial theorem. Since $f(x)=(1-2x)^{19}$ is a polynomial, it is equal to its Taylor series, that is, $$f(x) = \sum_{k=0}^\infty \frac{f^{(k)}(0)}{k!}x^k. $$ It's clear that for $0\leqslant k\leqslant 19$, $$f^{(k)}(x) = (-2)^k(19)_k(1-2x)^{19-k}, $$ where $(n)_k$ denotes the falling factorial: $$(n)_k = n(n-1)\cdots(n-k+1). $$ So $f^{(k)}(0) = (-2)^k(19)_k$ as $f^{(k)}=0$ for $k>19$, we have $$f(x) = \sum_{k=0}^{19} \frac{(-2)^k(19)_k}{k!}x^k. $$ It follows that the coefficient of $x^{12}$ is $$\frac{(-2)^{12}(19)_{12}}{12!}=2^{12}\binom {19}{12}. $$