Derivative Magic I

$\large{\begin{aligned}f(x)&=\displaystyle\sum_{k=1}^{n} \dfrac{x^{k+1}}{k}\\f'(x)&=\displaystyle\sum_{k=1}^{n} \dfrac{(k+1)x^{k}}{k}\\f''(x)&=\displaystyle\sum_{k=1}^{n} \dfrac{(k+1)\cancel{k}x^{k-1}}{\cancel{k}}\\f^3(x)&=\displaystyle\sum_{k=1}^{n}(k+1)( k-1)x^{k-2}\\&\cdots\\&\cdots\\ f^n~(x)&=\displaystyle\sum_{k=1}^{n}(k+1)(k-1)\cdots(k-(n-2))x^{(k-(n-1))}\end{aligned}}$

Since first $(n-2)$ terms of the summation will be zero we only have to calculate last two terms i.e:

$=\displaystyle\sum_{k=(n-1)}^{n}((k+1)(k-1)\cdots(k-(n-2))x^{(k-(n-1))}$

$\large{ =n(n-2)!+x(n+1)(n-1)!\\=\boxed{(n-2)!(n+x(n+1)(n-1))}}$

In the given question, $n=1000, x=100$ . Putting we get $(999)!\times (100201001)$ . Hence answer $=999+100201001=\boxed{100202000}$ .

We know that by deriving 1001 times, only the terms $\frac{x^{1001}}{1000}$ and $\frac{x^{1002}}{1001}$ will survive (all the rest become constants and get a derivative of zero).

Now deriving them 1001 times, we get :

$\frac{1001!}{1000} + \frac{1002!}{1001} . 100$ , where the 100 comes from the evaluation at $x = 100$ .

This simplifies to $999!.(1001 + 1002.100.1000)$ , which gives us 100202000.

Let us Apply Taylor's Theorem in the region of $(100 , 100+x)$ .
Let $f(x)$ = $\sum_{k=1}^{1001} \frac{x^{k+1}}{k}$

$f(100+x)$ = $f(100)$ + $\frac{xf'(100)}{1!}$ + $\frac{(x^2)f''(100)}{2!}$ +...... Now we just need to evaluate the coefficient of $x^{1001}$ in the above expression which is $\frac{1001+(1002).100000}{(1000)(1001)}$ . Now multiply it by $1001!$ to get $(100201001).(999!)$

We know nth derivative of x^n = constant = n! So only last 2 term of series are useful and other are zero so We have- 1001 999!+1002 1000!*100 Solving we get 999!(100201001) Add them to get 100202000 And please specify in question that a,b are integers.