<s>[INST] How to show $I_p(a,b) = \sum_{j=a}^{a+b-1}{a+b-1 \choose j} p^j(1-p)^{a+b-1-j}$ 
Show that $$I_p(a,b) = \frac{1}{B(a,b)}\int_0^p u^{a-1}(1-u)^{b-1}~du\\= \sum_{j=a}^{a+b-1}{a+b-1 \choose j} p^j(1-p)^{a+b-1-j}$$ when $a,b$ are positive integers.

I have no idea how to proceed. Please help.
 [/INST] Hint: Use $\frac{1}{B(a,b)}=\frac{\Gamma(a+b)}{\Gamma(a)\Gamma(b)}=\frac{(a+b-1)!}{(a-1)!(b-1)!}$ and integration by parts to evaluate the integral:
$$I=\int_{0}^{1}u^{a-1}(1-u)^{b-1} = \left.\frac{1}{a}u^a(1-u)^{b-1}\right|_{0}^{p}+\frac{b-1}{a}\int_{0}^{p}u^{a}(1-u)^{b-2}\,du, $$
$$ I= \frac{1}{a}p^a(1-p)^{b-1}+\frac{b-1}{a}\int_{0}^{p}u^{a}(1-u)^{b-2}\,du, $$
$$ I = \frac{1}{a}p^a(1-p)^{b-1} + \frac{b-1}{a(a+1)}p^{a+1}(1-p)^{b-2} +\frac{(b-1)(b-2)}{a(a+1)}\int_{0}^{p}u^{a+2}(1-u)^{b-3}\,du = \ldots $$
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