Question

S=1/1+2+1/1+2+3+1/1+2+3+4...+...
S<0.9

Answer 1

Răspuns:

19

Explicație pas cu pas:

$S = \frac{1}{1 + 2} + \frac{1}{1 + 2 + 3} + \frac{1}{1 + 2 + 3 + 4} + ... + \frac{1}{1 + 2 + ... + n} > 0.9$

$\frac{1}{ \frac{2 \times 3}{2} } + \frac{1}{ \frac{3 \times 4}{2} } + \frac{1}{ \frac{4 \times 5}{2} } + ... + \frac{1}{ \frac{n(n + 1)}{2} } > 0.9$

$\frac{2}{2 \times 3} + \frac{2}{3 \times 4} + \frac{2}{4 \times 5} + ... + \frac{2}{n(n + 1)} > \frac{9}{10}$

$2\Big( \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + \frac{1}{4} - \frac{1}{5} + ... + \frac{1}{n} - \frac{1}{n + 1} \Big) > \frac{9}{10}$

$2\Big( \frac{1}{2} - \frac{1}{n + 1} \Big) > \frac{9}{10} \iff \frac{2}{2} - \frac{2}{n + 1} > \frac{9}{10}$

$1 - \frac{9}{10} > \frac{2}{n + 1} \iff \frac{2}{n + 1} < \frac{1}{10} \\ \frac{1}{n + 1} < \frac{1}{20} \iff n + 1 > 20 \\ \implies \bf n > 19 \iff n \geqslant 20$

pentru n ≥ 20, S > 0,9 și cel mai mic număr de termeni este 19

$S = \frac{1}{1 + 2} + \frac{1}{1 + 2 + 3} + \frac{1}{1 + 2 + 3 + 4} + ... + \frac{1}{1 + 2 + ... + 20} > 0.9$

Answer 2

$\it 1+2+3+\ ...\ +n=\dfrac{n(n+1)}{2} \Rightarrow\dfrac{1}{1+2+3+\ ...\ +n}=\dfrac{2}{n(n+1)}\\ \\ \\ Suma\ devine:\\ \\ \\ S=2\Big(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\ ...\ +\dfrac{1}{n(n+1)}\Big)$

Observăm că suma conține n-1 termeni, deci trebuie să determinăm

cea mai mică valoare a lui n-1, pentru care S > 0,9 .

$\it Folosind\ formula\ \ \dfrac{1}{n(n+1)}=\dfrac{1}{n}-\dfrac{1}{n+1},\ \ \ suma\ \ devine:\\ \\ \\ S=2\Big(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\ ...\ +\dfrac{1}{n}-\dfrac{1}{n+1}\Big)=2\Big(\dfrac{^{n+1)}1}{\ \ 2}-\dfrac{^{2)}1}{n+1}\Big)=\\ \\ \\ =2\cdot\dfrac{n-1}{2(n+1)}=\dfrac{n-1}{n+1}$

$S > 0,9 \Rightarrow \dfrac{n-1}{n+1} > \dfrac{9}{10} \Rightarrow10(n-1) > 9(n+1) \Rightarrow 10(n-1) > 9[(n-1)+2] \Rightarrow \\ \\ \\ \Rightarrow 10(n-1) > 9(n-1)+18 \Rightarrow 10(n-1)-9(n-1) > 18 \Rightarrow n-1 > 18 \Rightarrow \\ \\ \\ \Rightarrow n-1\geq19$

Prin  urmare, cel mai mic număr de termeni ai sumei S,

pentru care S > 0,9, este egal cu 19.