Binomial series differential equation

Question

Show that

$$(1 + x) g'(x) = \ell g(x)$$

where $$g(x)=\sum _{k=0}^{\infty } x^k \binom{\ell}{k}$$

Answer 1

$$g(x)=\sum _{k=0}^{\infty } x^k \binom{\ell}{k}=(1+x)^{\ell}$$ $$g'(x)=\ell(1+x)^{\ell-1}$$ $$(1+x)g'(x)=\ell(1+x)(1+x)^{\ell-1}=\ell(1+x)^{\ell}=\ell g(x)$$