The $21^{\text{st}}$ Change

One way of solving this is to derive the expression $k^3 \sin k^2 \,21$ times and then evaluate at $k=0$ . That method kinda consumes quite some time though.

Another way, and I guess more efficient way of answering this is by using the concept of infinite series, Maclaurin Series in particular.

Maclaurin Series is a special type of Taylor Series centered at $a=0$ : $\\ \quad \qquad \qquad \qquad \qquad \qquad f(x)=\displaystyle \sum_{n=0}^\infty \frac{f^n(0)}{n!}x^n$

The first step then is to write the function we are about to analyze as an infinite series. A list of common Maclaurin series can be found in this list of mathworld.wolfram.com. We see that: $\\ \quad \qquad \qquad \qquad \qquad \qquad \sin k=\displaystyle \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}k^{2n+1}$

where the expression $(-1)^n$ corresponds to the $(2n+1)^{th}$ derivative of $\sin k$ at $k=0$ . Using this known relation, we can then rewrite $k^3 \sin k^2$ as an infinite series: $\\ \quad \qquad \qquad \qquad \qquad \qquad k^3 \sin k^2=k^3 \displaystyle \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}(k^2)^{2n+1}$ $\\ \quad \qquad \qquad \qquad \qquad \qquad k^3 \sin k^2=\displaystyle \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}k^{4n+5}$

We then rewrite the above series so similar to that of the expression for a Maclaurin series $f(x)$ . $\\ \quad \qquad \qquad \qquad \qquad \qquad k^3 \sin k^2=\displaystyle \sum_{n=0}^\infty \frac{\frac{(-1)^n(4n+5)!}{(2n+1)!}}{(4n+5)!}k^{4n+5}$

where $\frac{(-1)^n(4n+5)!}{(2n+1)!}$ is the $(4n+5)th$ derivative of the expression evaluated at $k=0$ and hence $N(0)=g(n)=\frac{(-1)^n(4n+5)!}{(2n+1)!}$ . Since we want to determine the $21^{st}$ derivative of the expression, we let $4n+5=21$ and see that $n=4$ . Therefore, $N(0)=g(4)=\frac{21!}{9}=140792940288000$ and the sum of its digits is $54$ .