chứng minh ..........

Question

Cho $A=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{16}+\frac{1}{17}$ .

Chứng minh rằng : $A$ không phải là số tự nhiên.

tran85295 · Answer 1 · 3/23/2016 6:56:43 PM

Ta có

$\frac1{17} < \frac1{16},\frac1{15}< \frac 1{14},...$

$\Rightarrow \frac 19+\frac 1{11}+\frac 1{13}+\frac{1}{15}+ \frac{1}{17} < \frac 1{8} +\frac 1{10} + \frac 1{12} + \frac 1{14}+ \frac 1{16}$

$\Rightarrow \frac 18+\frac 19+\frac 1{10}+...+\frac 1{17}<2 (\frac 1{8} +\frac 1{10} + \frac 1{12} + \frac 1{14}+ \frac 1{16})$

$\Rightarrow \frac 18+\frac 19+\frac 1{10}+...+\frac 1{17} < \frac 14+\frac 15 + \frac 16 + \frac 17 +\frac 18$

$\Rightarrow \frac 14+ \frac 15 +\frac 16+...+ \frac 1{17}<2(\frac 14+\frac 15+\frac 16+ \frac 17)+\frac 18<2.1 +\frac 18$

$\Rightarrow A< \frac 12+\frac 13+2+ \frac 18<3$

Lại có

$A> \frac 12+\frac 13+\frac 14+\frac 15+\frac 16 + \frac 17 + \frac 18+\frac 1{10}+ \frac 1{12}+\frac 1{14} +\frac 1{15} + \frac 1{16}$

$=(\frac 12+ \frac 13+ \frac 1{10} + \frac 1{15})+(\frac 15 +\frac 17)+ (\frac 1{14}+ \frac{1}{16})+( \frac 14+\frac 16 + \frac 18+ \frac 1{12})$

$>1+(\frac 18 + \frac 18)+(\frac 1{16} + \frac 1{16})+ \frac 58 =2$

Vậy ta có

$2<A<3$ nên

$A$ ko là số tự nhiên

score 0 · Answer 2

a=13/17

sai rồi :D – ๖ۣۜJinღ๖ۣۜKaido 23-03-16 05:14 PM

sao ngắn gọn vậy ? – Hàn Thiên Dii 23-03-16 04:12 PM

2 Đáp án