Find the 66th derivative of this integral function. $$F\left(x\right)=\int _0^x\cos\left(t^3\right)dt$$. 
Writing in the form of an infinite series, we get $$\sum _{n\ge 1}\left(\frac{\left(-1\right)^n}{\left(2n\right)!}\cdot \frac{x^{6n+1}}{6n+1}\right)$$
How can we find $F^{\left(66\right)}\left(0\right)$?
Here's one solution that comes with this exercise, but which I can't comprehend:
$$6n+1=66$$ $$n=\frac{65}{6}\notin \mathbb{N}.$$ Therefore $F^{\left(66\right)}\left(0\right)=0$.
 A: The coefficients of a power series are related to the derivatives of the function:
$$F(x)=\sum_{k=0}^\infty \frac{F^{(k)}(0)}{k!}x^k.$$
Thus the coefficient of $x^{66}$ will tell us about the 66th derivative of $F(x)$. However, $F$ does not have an $x^{66}$ term in its power series expansion, since every term is of the form $x^{6n+1}$ for some integer $n$. Thus $\frac{F^{(66)}(0)}{66!}=0$.
A: Since $t\mapsto\cos(t^3)$ is an even function, $F$ is an odd function and therefore $F^{(n)}(0)=0$ for every even number $n$.
A: One important point to note here is that whenever you see a problem some peculiarly large number put in place that otherwise would make the problem difficult to tract, that's meant to flag that there's a trick you have to look for, which makes the problem much easier.
In this case, the trick is this. The cosine series contains only even powers, so the series for the function $f(x) := \cos(x^3)$ now contains only powers that are multiples of six ($2 \cdot 3$). Or equivalently, the coefficient of any $x^n$ in the series of $f$ for which $n$ is not a multiple of 6 is zero. Hence, say, the coefficient of $x^{1000000000}$ in the series for $f$ is 0: we can surmise this instantaneously without any sort of intractable calculation because no power of 10 is divisible by 6 (since 6 contains a prime factor 3, but 10 does not).
Likewise, in this case, you are trying to do the same. The integral, in effect, shifts this pattern of zero and nonzero coefficients "one up": instead of it now being that all coefficients are zero except for those whose indices $n$ have the form $n = 6k$ for some integer $k$, it is now that they are all zero except for those with orders $n$ of the form $n = 6k + 1$.
And $66 = 6 \cdot 11$, which is not of the form $6k + 1$, but instead of $6k$. Hence its coefficient in the integral series must be zero, and so $F^{(66)}(0) = 0$.
A: The term-by-term derivative formula implies that the 66-th derivative is
$$\sum _{n\ge *}\left(\frac{\left(-1\right)^n}{\left(2n\right)!}\cdot \frac{x^{6n+1-66}}{6n+1}\cdot(6n+1)(6n+1-1)(6n+1-2)\cdot(6n+1-65)\right)$$
The * at the bottom of the summation means that all the terms whose exponents would be negative above should disappear.
Then when evaluated at $x=0$, all the terms having positive exponents would also disappear.  So the only terms remaining would have exponent 0.
Then the "solution" you quoted implies that there is no $n$ for which the exponent would be $0$.  This means that there are no terms remaining, period, so the coefficient is the sum of 0 terms, so equal to 0.
