Question

Buna seara! Va rog sa ma ajutati cu aceasta suma.
Sigma 4k/(4k la puterea 4 + 1)

Va multumesc!

Answer 1

Răspuns:

$4k^4+1=4k^4+4k^2+1-4k^2=(2k^2+1)^2-(2k)^2=\\=(2k^2-2k+1)(2k^2+2k+1)$

Atunci

$\dfrac{4k}{4k^4+1}=\dfrac{1}{2k^2-2k+1}-\dfrac{1}{2k^2+2k+1}=\dfrac{1}{2(k-1)k+1}-\dfrac{1}{2k(k+1)}}$

Rezultă

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

Explicație pas cu pas: