Vì \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{n\left(n+1\right)}=1-\frac{1}{n+1}< 1\)=> Q < 1 (đpcm)
Chứng minh:
a. \(A=\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}< \dfrac{1}{3}\)
b.\(B=\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}< \dfrac{3}{16}\)
c. \(C=\dfrac{1}{41}+\dfrac{1}{42}+\dfrac{1}{43}+...+\dfrac{1}{79}+\dfrac{1}{80}>\dfrac{7}{12}\)
1. Chứng tỏ :
a. 1/31+1/32+..+1/60 không thuộc Z
giup minh nhé các bạn!
Chứng minh rằng 1/1.2 + 1/2.3 + 1/3.4 +........+1/49+50 = 1/26 + 1/27 +1/28 +.....+ 1/50
cho S =1/5^2+1/5^4+1/5^6+...+1/5^2014 chứng minh S<1/24
1)Tinh 9/10-1/90-1/72-1/56-1/42-1/30-1/20-1/12-1/6-1/2
Chứng minh (x-1)(x^2+x+1)=x^3-1
Chứng minh rằng: \(\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+....+\dfrac{1}{\sqrt{100}}>10\)
Cho A= 1/1.2 +1/2.3 +1/3.4 +......+1/49+50. Chứng minh rằng 7/ 12 < A < 5/6
\(\frac{\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{199.200}}{\frac{1}{101}+\frac{1}{102}+...\frac{1}{200}}=1\)
Hãy chứng minh
Cho :A=1/1.2 +1/2.3 + 1/3.4 + .......+1/49+50. Chứng minh rằng 7/12 < A < 5/6.
