Show that Newton’s Binomial series for any exponent $p$ will be equal to the corresponding function by verifying the differential equation:$$(1 +x)y′=py$$You will likely need the generalized “Pascal triangle rule”
$${p \choose k}={p−1 \choose k−1}+{p−1 \choose k}$$
I have that Newton's Binomial series says that
$$(1+x)^p=1+px+\frac{p(p−1)}{2!}x^2+\frac{p(p−1)(p−2)}{3!}x^3+···$$
I have also that
$$\begin{align*} (1 +x)y′=py &\iff\frac{1}{py}\frac{d}{dx}(y)=\frac{1}{1+x}\\\\ &\iff\int\frac{1}{py}dy=\int\frac{1}{1+x}dx\\\\ &\iff\frac{1}{p}ln(y)+c_1=ln(1+x)+c_2\\\\ &\iff ln(y)=p\cdot ln(1+x)+pc_3\\\\ &\iff ln(y)=ln\left(e^{pc_3}(x+1)^p\right)\\\\ &\iff y=e^{pc_3}(x+1)^p\\\\ &\iff y=c_4 (x+1)^p \end{align*}$$
which equals $(1+x)^p$ if $c_4=1$ but I don't think this is what I am supposed to show, since I did not make use of the given Pascal triangle rule hint.
I think I need to use the given differential equation to show
$$1+px+\frac{p(p−1)}{2!}x^2+\frac{p(p−1)(p−2)}{3!}x^3+···$$
but I'm not sure how to go about doing that.
Any hints would be much appreciated.
Edit:
Differentiating the RHS of
$$(1+x)^p=1+px+\frac{p(p−1)}{2!}x^2+\frac{p(p−1)(p−2)}{3!}x^3+···$$
I get
$$p+2\frac{p(p-1)}{2!}x+3\frac{p(p-1)(p-2)}{3!}x^2+...$$ $$= p+2{p \choose 2}x+3{p \choose 3}x^2+...\\\\$$ $$= p+2\left({p-1 \choose 2-1}+{p-1 \choose 2}\right)x+3\left({p-1 \choose 3-1}+{p-1 \choose 3}\right)x^2+...$$
but I'm not sure how to proceed from here.
You did half. Now you are supposed to show that the series satisfies the same differential equation, which you can do by differentiating term by term and applying Pascal's triangle...