Newtons Binomial Theorem identity

Question

I have seen a lot of identites being discussed here but I still haven't seen the one I'm having a problem with.

I need to conclude that $\sum\limits_{i=1}^{n} i\binom{n}{i} \frac{(-4)^{i-1}}{5^{i-n}} = n$

I'm supposed to derive something from Newtons binomial theorem but I can't see a starting point. Whats so special about $\frac{-4}{5}$ and their superscripts. I see that below the fractal line 5 will always be a small number and -4 will keep getting bigger, meaning the overall number gets bigger and iterates between odd/even = positive/negative, do they cancel out in the end?

Where should I start from?

Best regards

Answer 1

Differentiate the identity $$ (1+x)^n=\sum_{i=0}^n\binom{n}{i} x^i $$ to wind up with $$ n(1+x)^{n-1}=\sum_{i=1}^ni\binom{n}{i}x^{i-1}\tag{0}. $$ Observe that your sum can be written as $$ \sum_{i=1}^{n} i\binom{n}{i} \frac{(-4)^{i-1}}{5^{i-n}} =\frac{1}{5^{1-n}}\sum_{i=1}^ni\binom{n}{i}\left(\frac{-4}{5}\right)^{i-1} $$ and use equation $(0)$.

Answer 2

Consider: $$ f(x)=(5-4x)^n=\sum_{i=0}^n\binom ni (-4x)^i 5^{n-i}. $$

Then: $$ \frac{df(x)}{dx}=\sum_{i=0}^n\binom ni i(-4x)^{i-1} 5^{n-i}. $$

Thus: $$ \sum_{i=0}^n\binom ni i(-4x)^{i-1} 5^{n-i}=\lim_{x\to1}\frac{d (5-4x)^n}{dx} =\lim_{x\to1} n(5-4x)^{n-1}=n. $$