Let n be an element of the set of natural numbers
Let $F(x)=x^2(1+x)^n$ and write $F^n$ for the nth derivative of the function $F$.
Compute $F^n$ by applying the Binomial Theorem to $(1+x)^n$.
I don't understand the step where I need to find the derivative of these terms:
= $x^2 + {n \choose 1}x^3 + {n \choose 2} x^4 + …. + {n \choose n-1} x^{n+1} + x^{n+2} $
Even though I do understand how to show that for the first 3 terms the nth derivative is 0, but for the last two I have no clue??
Really appreciate any hints!!
Best Answer
The $n$-th derivative is $0$ for many more than just the first three terms: in fact, for all but the last three.
You have a typo, the last term is $x^{n+2}$.
The last three terms are $${n\choose{n-2}} x^n+{n\choose{n-1}} x^{n+1}+x^{n+2}$$
Can you work it out from here? Some factorials will appear when you take the $n$-th derivative, hence you might want to express the $n\choose k$ in terms of factorials as well, to see if something cancels out, and at the very least to get a nice-looking formula.