Question

va rooooooog e urgent​

Answer 1

$\it \dfrac{^{n+2)}1}{\ \ n}-\dfrac{^{n)}1}{n+2}=\dfrac{n+2-n}{n(n+2)}=\dfrac{2}{n(n+2)}$

Folosind formula de mai sus, expresia de calculat devine:

$\it \dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}=\dfrac{^{11)}1}{\ 3}-\dfrac{^{3)}1}{\ 11}=\dfrac{11-3}{33}=\dfrac{8}{33}$