Compute from Left-Side: $$ {2p \choose p} -{2p \choose p-1} = {(\frac {1}{p+1})} {2p \choose p}$$
This is the answer
$$ ={2p \choose p} -{2p \choose p - 1}$$ $$=\left(\frac{(2p!)}{(p!) (p!)}\right)-\left(\frac{(2p)!}{(p-1)! (p+1)!}\right) $$ $$=\left(\frac{(2p!)}{(p!) (p!)}\right) \left(1-\frac{p}{p+1}\right) $$ $$={2p \choose p} {\left(\frac {p+1-p}{p+1}\right)} $$ $$={\frac {1}{p+1}} {2p \choose p}$$ $$=RS$$ I'm really trying to understand the algebraic part of this example. Could someone me help understand this problem by breaking it step-by-step?
Especially this part:
$$=\left(\frac{(2p!)}{(p!) (p!)}\right) \left(1-\frac{p}{p+1}\right) $$ $$...$$ $$...$$ $$...$$ Thanks
$$ ={2p \choose p} -{2p \choose p - 1}$$ Apply definition of $^nC_r=\frac{n!}{r!(n-r)!}$. $$=\left(\frac{(2p!)}{(p!) (p!)}\right)-\left(\frac{(2p)!}{(p-1)! (p+1)!}\right) $$ Extra step. Rewrite $(p-1)!$ as $\frac{p!}{p}$ and $(p+1)!$ as $p!(p+1)$.
$$=\left(\frac{(2p!)}{(p!) (p!)}\right)-\left(\frac{p(2p)!}{p! p!(p+1)}\right) $$ Take out common factor. $$=\left(\frac{(2p!)}{(p!) (p!)}\right)\left(1-\frac{p}{p+1}\right) $$ Find common denominator $$={2p \choose p} {\left(\frac {p+1-p}{p+1}\right)} $$ Simplify $$={\frac {1}{p+1}} {2p \choose p}$$ Done $$=RS$$