Question

Va rog !!! Dau coroana !!!​

Answer 1

termenul general este:

$T_{n} = \sqrt{n^{2} + n + 1} - \sqrt{n^{2} - n + 1})$

$n = 1 \iff \sqrt{1^{2} + 1 + 1} - \sqrt{1^{2} - 1 + 1} = \sqrt{3} - \sqrt{1}$

$n = 2 \iff \sqrt{2^{2} + 2 + 1} - \sqrt{2^{2} - 2 + 1} = \sqrt{7} - \sqrt{3}$

$n = 3 \iff \sqrt{3^{2} + 3 + 1} - \sqrt{3^{2} - 3 + 1} = \sqrt{13} - \sqrt{7}$

...

$\sqrt{(n - 1)^{2} + (n - 1) + 1} - \sqrt{(n - 1)^{2} - (n - 1) + 1} = \sqrt{n^{2} - 2n + 1 + n - 1 + 1} - \sqrt{n^{2} - 2n + 1 - n + 1 + 1} = \sqrt{n^{2} - n + 1} - \sqrt{n^{2} - 3n + 3}$

suma se scrie:

$S_{n} = (\sqrt{3} - \sqrt{1}) + (\sqrt{7} - \sqrt{3}) + (\sqrt{13} - \sqrt{7}) + ... + (\sqrt{n^{2} - n + 1} - \sqrt{n^{2} - 3n + 3}) + (\sqrt{n^{2} + n + 1} - \sqrt{n^{2} - n + 1})$

anulăm termenii asemenea:

$S_{n} = - \sqrt{1} + \sqrt{n^{2} + n + 1} =\bf \sqrt{n^{2} + n + 1} - 1$