Q. If $$f(x) =\displaystyle \prod_{r=0}^{n} \left( \binom{n}{r} x + r+1 \right)$$ and $$f'(0)= \frac{3}{2} (7!)$$ then find the value of n.
My Attempt- I took $\ln$ both sides to get- $$\ln(f(x)) = \displaystyle\sum_{r=0}^{n}\left(\ln\left(\binom{n}{r} \, x +r+1\right) \right)$$
Now, If we differentiate $\ln(f(x))$ and put $x=0$, we get - $$ \frac{f'(0)}{f(0)} = \frac{f'(0)}{(n+1)!}= \displaystyle\sum_{r=0}^{n} (\frac{\binom{n}{r}}{r+1})$$
Now, I'm stuck and clueless from here on!
Any help would be appreciated
Yes, you are correct, $$f'(x)=f(x)\sum_{r=0}^n\frac{\binom{n}{r}}{\binom{n}{r} x + r+1}\implies f'(0)=(n+1)!\sum_{r=0}^n\frac{\binom{n}{r}}{r+1}.$$ Now consider $$\int_0^1(x+1)^ndx=\sum_{r=0}^n\binom{n}{r}\int_0^1x^rdx.$$ Can you take it from here?