Cho \(A=\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{n}{5^{n+1}}+...+\frac{11}{5^{12}}\)
CM rằng A \(\frac{1}{16}\)
GIÚP MK VS
AI LÀM ĐÚNG MK TICK Ạ
Cho A = \(\frac{1}{5^2}+\frac{1}{5^3}+\frac{1}{5^4}+.......+\frac{n}{5^{n+1}}+.......+\frac{11}{5^{12}}\) với n \(\in\) N. Chứng minh rằng A < \(\frac{1}{16}\)
Giúp mk vs
Cho \(A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}....\frac{199}{200}\)
Chứng minh rằng A2 <\(\frac{1}{201}\)
Chứng minh rằng luôn luôn tồn tại số tự nhiên n để
\(1+\frac{1}{2}+\frac{1}{3}+...\frac{1}{n}>1000\)
Mn giúp mk vs ạ
mk đg cần gấp
Ai làm đc ,dúng mk tick cho !!~~~
Cho A = \(\frac{1}{5^2}+\frac{2}{5^3}+\frac{3}{5^4}+........+\frac{n}{5^{n+1}}+............+\frac{11}{5^{12}}\)
với n \(\in N\). Chứng minh rằng A < \(\frac{1}{16}\)
ai giải đc mình tick 2
Cho A=\(\frac{1}{5^2}+\frac{2}{5^3}+\frac{3}{5^4}+...+\frac{n}{5^{n+1}}+...+\frac{11}{5^{12}}\)với n\(\inℕ\).Chứng minh rằng A<\(\frac{1}{16}\)
Giúp mình với, hiện đang cần gấp lắm.
5A=\(\frac{1}{5}+\frac{2}{5^2}...+\frac{n}{5^n}...+\frac{11}{5^{11}}\)
=>4A=5A-A=\(\frac{1}{5}+\frac{1}{5^2}...+\frac{1}{5^{11}}-\frac{11}{5^{12}}\)
=>20A=\(1+\frac{1}{5}+...+\frac{1}{5^{10}}-\frac{11}{5^{11}}\)
=>16A=20A-4A=\(1-\frac{1}{5^{11}}+\frac{11}{5^{12}}-\frac{11}{5^{11}}\)
Mà \(1-\frac{1}{5^{11}}< 1\),\(\frac{11}{5^{12}}-\frac{11}{5^{11}}< 0\)
=>16A<1
Do đó: A<1/16(đpcm)
\(1/ Cho P=\frac{1}{5^2}+\frac{2}{5^3}+\frac{3}{5^4}+\frac{4}{5^5}+...+\frac{11}{5^{12}}.\) Chứng minh rằng \(P<\frac{1}{16}\)
\(2/ \) \(Cho\) \(A=2009^{2010}+2010^{2010}+2011^{2010}\). Số A có là số chính phương không?
Ai nhanh mk tick nha
1) \(P=\frac{1}{5^2}+\frac{2}{5^3}+\frac{3}{5^4}+...+\frac{11}{5^{12}}\)
\(5P=\frac{1}{5^1}+\frac{2}{5^2}+\frac{3}{5^3}+...+\frac{11}{5^{11}}\)
\(5P-P=\frac{1}{5^1}+\left(\frac{2}{5^2}-\frac{1}{5^2}\right)+\left(\frac{3}{5^3}-\frac{2}{5^3}\right)+...+\left(\frac{11}{5^{11}}-\frac{10}{5^{11}}\right)-\frac{11}{5^{12}}\)
\(4P=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{11}}-\frac{11}{5^{12}}\)
Đặt \(A=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{11}}\)
\(5A=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{10}}\)
\(5A-A=1+\frac{1}{5}-\frac{1}{5}+\frac{1}{5^2}-\frac{1}{5^2}+...+\frac{1}{5^{10}}-\frac{1}{5^{11}}\)
\(4A=1-\frac{1}{5^{11}}\Rightarrow A=\frac{1-\frac{1}{5^{11}}}{4}\)
\(4P=\frac{1-\frac{1}{5^{11}}}{4}-\frac{11}{5^{12}}=\frac{1-\frac{1}{5^{11}}}{16}-\frac{11}{5^{12}\cdot4}< \frac{1}{16}\)
Cho A = \(\frac{1}{5^2}\)+ \(\frac{2}{5^3}\)+ \(\frac{3}{5^4}\)+ ..... + \(\frac{n}{5^{n+1}}\)+ ...... + \(\frac{11}{5^{12}}\)với n thuộc N. CMR: A < \(\frac{1}{16}\)
Giúp mk với nhé!
tính
1/\(\frac{\left(\frac{3}{10}-\frac{4}{15}-\frac{7}{20}\right).\frac{5}{9}}{\left(\frac{1}{4}+\frac{1}{7}-\frac{-3}{35}\right).\left(-1\frac{1}{3}\right)}\)
2/\(\frac{0,6-\frac{1}{3}+\frac{3}{11}}{1,4-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-0,25+\frac{1}{5}}{1\frac{1}{6}-0,875+0,7}\)
các bn lm giúp mk vs!!!!
mk đang cần gấp lắm nha!!!!!!!!!
ai lm nhanh nhất đúng đủ trình bày khoa học mk tick cho !!!!!!!1
a) Cho A= \(\left|x-\frac{1}{3}\right|+\frac{1}{4}\)
So sánh A với \(\frac{1}{5}\)
giúp mk vs, ai lm nhanh nhất đúng nhất mk tick cho
\(A=\left|x-\frac{1}{3}\right|+\frac{1}{4}\ge\frac{1}{4}>\frac{1}{5}\)
TÍNH : H= \(\frac{1}{5^2}\)+\(\frac{2}{5^3}\)+........+\(\frac{11}{5^{12}}\)
hic. giúp mk vs T .T .
À thêm 1 tí nhé : Cm H < \(\frac{1}{16}\)
\(\Rightarrow5H=\frac{1}{5}+\frac{2}{5^2}+...+\frac{11}{5^{11}}\)
\(\Rightarrow5H-H=\left(\frac{1}{5}+\frac{2}{5^2}+...+\frac{11}{5^{11}}\right)-\left(\frac{1}{5^2}+\frac{2}{5^3}+...+\frac{11}{5^{12}}\right)\)
\(\Rightarrow4H=\frac{1}{5}+\frac{1}{5^2}+..+\frac{1}{5^{11}}-\frac{11}{5^{12}}\)
Đặt \(A=\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{11}}\)
\(\Rightarrow5A=1+\frac{1}{5}+...+\frac{1}{5^{10}}\)
\(\Rightarrow5A-A=\left(1+..+\frac{1}{5^{10}}\right)-\left(\frac{1}{5}+...+\frac{1}{5^{11}}\right)\)
\(\Rightarrow4A=1-\frac{1}{5^{11}}\)
\(\Rightarrow A=\frac{1}{4}-\frac{1}{4.5^{11}}\)
\(\Rightarrow4H=\frac{1}{4}-\frac{1}{4.5^{11}}-\frac{11}{5^{12}}\)
\(\Rightarrow H=\frac{1}{16}-\frac{1}{4^2.5^{11}}-\frac{11}{4.5^{12}}\)
Ta có : \(5H=\frac{1}{5}+\frac{2}{5^2}+...+\frac{11}{5^{11}}\)
\(\Rightarrow4H=\left(\frac{1}{5}+\frac{2}{5^2}+...+\frac{11}{5^{11}}\right)-\left(\frac{1}{5^2}+\frac{2}{5^3}+...+\frac{11}{5^{12}}\right)=\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{11}}+\frac{11}{5^{12}}\)
\(\Rightarrow20H=1+\frac{1}{5}+...+\frac{1}{5^{10}}+\frac{11}{5^{11}}\)
\(\Rightarrow16H=20H-4H=1+\frac{10}{5^{11}}-\frac{11}{5^{12}}\Leftrightarrow H=\frac{1+\frac{10}{5^{11}}-\frac{11}{5^{12}}}{16}.\)
Đỗ Đức Lợi nhầm rồi
Phải là \(\frac{1+\frac{1}{5^{11}}-\frac{11}{5^{12}}}{16}\)nhé cu