Question

ma puteti ajuta ???????

Answer 1

$\text{Fie:}\\\\E(a,b) = (a+1)\Big(\dfrac{a}{2}+1\Big)\Big(\dfrac{a}{3}+1\Big) ... \Big(\dfrac{a}{b}+1\Big) \\ \\= \Big(\dfrac{a+1}{1}\Big)\Big(\dfrac{a+2}{2}\Big)\Big(\dfrac{a+3}{3}\Big)...\Big(\dfrac{a+b}{b}\Big)\\ \\ =\dfrac{(a+1)(a+2)(a+3)...(a+b)}{1\cdot 2\cdot 3\cdot ...\cdot b}\\ \\ = \dfrac{(a+1)(a+2)(a+3)...(a+b)}{b!}\\ \\ = \dfrac{\dfrac{1\cdot 2\cdot 3\cdot...\cdot a\cdot(a+1)\cdot (a+2)\cdot ...\cdot (a+b)}{1\cdot 2\cdot 3\cdot ...\cdot a}}{b!} \\ \\ = \dfrac{\dfrac{(a+b)!}{a!}}{b!}$

$=\dfrac{(a+b)!}{a!\cdot b!} = \dfrac{(b+a)!}{b!\cdot a!}\\ \\ \\\text{Observam ca expresia este simetrica.}\\ \\ \Rightarrow E(a,b) = E(b,a) \\ \\\Rightarrow (a+1)\Big(\dfrac{a}{2}+1\Big)\Big(\dfrac{a}{3}+1\Big) ... \Big(\dfrac{a}{b}+1\Big) =\\ \\ = (b+1)\Big(\dfrac{b}{2}+1\Big)\Big(\dfrac{b}{3}+1\Big) ... \Big(\dfrac{b}{a}+1\Big)$