$\sqrt{1+x}$ as summation

91 Views Asked by At

I was reading this question and there was something in the first answer (adfriedman's answer) that I didn't understand.

It states that: $$ \int_0^1(1+(-x^2))^{1/2}dx = \int_0^1\sum_{k=0}^\infty\binom{1/2}{k}(-x^2)^k dx $$ Which I assume gives: $$ (1+(-x^2))^{1/2} = \sum_{k=0}^\infty\binom{1/2}{k}(-x^2)^k $$ I have never seen this, and I couldn't find anything about it online, so I was wondering where this comes from. Could it possibly be generalised to: $$ (1+a)^{b} = \sum_{k=0}^\infty\binom{b}{k}(a)^k $$

If so, where does this come from? What's the intuition behind it? Why does it make sense?