Doar pe C) .............

[tex]c)A(\left(\dfrac{1}{1\cdot 2}\right)\cdot A\left(\dfrac{1}{2\cdot 3}\right)\cdot \ldtos \cdot A\left( \dfrac{1}{2016\cdot 2017}\right)=A\left(\dfrac{a}{a+1}\right)\\\text{Folosind punctul anterior putem scrie:}\\A\left(\dfrac{1}{1\cdot 2}+\dfrac{1}{2\cdot 3}+\dfrac{1}{3\cdot 4}+\ldots +\dfrac{1}{2016\cdot 2017}\right)=A\left(\dfrac{a}{a+1}\right)\\\text{Hai sa ocupam de suma aia:}\\\dfrac{1}{1\cdot 2}+\dfrac{1}{2\cdot 3}+\ldots +\dfrac{1}{2016\cdot 2017}=[/tex]

[tex]1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\ldots +\dfrac{1}{2016}-\dfrac{1}{2017}\stackrel{\text{termenii se reduc }}{========} 1-\dfrac{1}{2017}=\dfrac{2016}{2017}\\\text{Revenind:}\\A\left(\dfrac{2016}{2017}\right)=A\left( \dfrac{a}{a+1}\right)\\\text{Se observa ca }a=2016.[/tex]

A(1 /1·2) +A(1 /2·3) +...+A(1 /2016·2017) =A(a /a+1) <=>

A(1 /1·2 +1 /2·3 +...+1 /2016·2017) =A(a /a+1) <=>

A(2016/2017)=A(a /a+1) => a /a+1 =2016/2017 => 2017a=2016a +1 => a=1 .

Observatie: pentru calcularea sumei din membrul stang al egalitatii se foloseste formula

1 /n·(n+k)=1/k ·(1/n -1 /(n+k)) ,unde 0 < k < n a.i. n;k∈N\{0} . In cazul nostru ,k=1 => 1 /n·(n+1)=1/n -1 /n+1 .