Question

Ex 30 multumesc(troll =spamm de reporturi)

Answer 1

Explicație pas cu pas:

formula termenului general:

$T_{n} = \dfrac{1}{ \sqrt{n + 1} + \sqrt{n} } = \dfrac{\sqrt{n + 1} - \sqrt{n}}{ (\sqrt{n + 1} + \sqrt{n})(\sqrt{n + 1} - \sqrt{n}) } =$

$= \dfrac{\sqrt{n + 1} - \sqrt{n}}{n + 1 - n} = \sqrt{n + 1} - \sqrt{n}$

atunci:

$S = \sqrt{2} - 1 + \sqrt{3} - \sqrt{2} + \sqrt{4} - \sqrt{3} + ... + \sqrt{n} - \sqrt{n - 1} + \sqrt{n + 1} - \sqrt{n}$

(se reduc termenii asemenea)

$S = - 1 + \sqrt{n + 1} = \sqrt{n + 1} - 1$

a)

$S = \sqrt{99 + 1} - 1 = \sqrt{100} - 1 = 10 - 1 = 9$

b)

$S = 50 \implies \sqrt{n + 1} - 1 = 50$

$\sqrt{n + 1} = 51 \iff n + 1 = {51}^{2}$

$n = 2601 - 1 \implies n = 2600$

c)

$S < 50 \implies \sqrt{n + 1} - 1 < 50 \iff \sqrt{n + 1} < 51$

$S \in \mathbb{N} \implies 0 \leqslant \sqrt{n + 1} \leqslant 50$

$n \in \mathbb{N} \implies 1 \leqslant \sqrt{n + 1} \leqslant 50$

$S = p, \ \ p \in \{1;2;3;...;49;50\}$

$\sqrt{n + 1} = p \implies n = {p}^{2} - 1$

=>

$p = 1 \implies n = {1}^{2} - 1 = 0$

$p = 2 \implies n = {2}^{2} - 1 = 3$

$p = 3 \implies n = {3}^{2} - 1 = 8$

...

$p = 49 \implies n = {49}^{2} - 1 = 2400$

$p = 50 \implies n = {50}^{2} - 1 = 2499$