Question

am incercat sa iau fiecare p si sa il integrez in formula ultimului termen dar de aici nu stiu ce sa mai fac... trebuie sa compar fiecare termen cu 28/15?

Answer 1

Răspuns:

Explicație pas cu pas:

Hai sa aducem sirul la o forma mai simpla.

$\displaystyle\dfrac{k^2}{(2k-1)(2k+1)}=\dfrac{1}{4}\cdot \dfrac{4k^2-1+1}{(2k-1)(2k+1)}=\dfrac{1}{4}\cdot\dfrac{4k^2-1}{(2k-1)(2k+1)}+\\\\\dfrac{1}{4}\cdot \dfrac{1}{(2k-1)(2k+1)}=\dfrac{1}{4}\cdot \dfrac{(2k-1)(2k+1)}{(2k-1)(2k+1)}+\dfrac{1}{8}\cdot \dfrac{2k+1-(2k-1)}{(2k-1)(2k+1)}=\\\\=\dfrac{1}{4}+\dfrac{1}{8}\left(\dfrac{1}{2k-1}-\dfrac{1}{2k+1}\right)\\\texttt{Prin urmare, suma devine:}\\a_n=\sum_{k=1}^n\left(\dfrac{1}{4}+\dfrac{1}{8}\left(\dfrac{1}{2k-1}-\dfrac{1}{2k+1}\right)\right)$

$\displaystyle a_n=\dfrac{n}{4}+\dfrac{1}{8}\left(\sum_{k=1}^n\dfrac{1}{2k-1}-\dfrac{1}{2k+1}\right)\\a_n=\dfrac{n}{4}+\dfrac{1}{8}\left(1-\dfrac{1}{2n+1}\right)\\\\a_n=\dfrac{n}{4}+\dfrac{1}{8}\cdot\dfrac{2n}{2n+1}\\\\a_n=\dfrac{n}{4}+\dfrac{n}{4(2n+1)}\\\\a_n=\dfrac{n(2n+1)+n}{4(2n+1)}\\\\a_n=\dfrac{n(2n+2)}{4(2n+1)}\\\\\boxed{a_n=\dfrac{n(n+1)}{2(2n+1)}}$

Sper ca te descurci singur mai departe.