Binomial Expansion - Simple application of the formula I found this one question in an old book:
What is the coefficient of $x^{n+1}$ in the expansion of $(x+2)^n \cdot x^3$?

Answer (according to my book): $(n^2-n) \cdot 2^{n-3}$

Here is my work: Since $\ T_{k+1} = \binom{n}{k}\cdot a^k\cdot b^{n-k}$, we can obtain a general-term for the expansion of $(x+2)^n \cdot x^3$. Letting $a = 2$ and $b = x$, then $ T_{k+1} = (\binom{n}{k}\cdot 2^k\cdot x^{n-k})\cdot x^3$.
So, we can easily notice that $x^{n+1}$ will show-up in the third term of the expansion (first we have $x^{n+3}$, then $x^{n+2}$ and $x^{n+1}$).
But, if we plug-in $k=2$ in $T_{k+1}$, its coefficient will be equal to $(n-3)! \cdot 2$.
Where is my mistake?

The binomial theorem yields
$$
(x+2)^n = \sum_{k=0}^n \binom nk x^k 2^{n-k},
$$
and so the coefficient of $x^{n+1}$ in $(x+2)^nx^3$ is
$$
\binom n{n-2}2^{n-(n-2)}) = \binom n2 2^2 = 2n(n-1).
$$
