Chứng minh: d=1/2^2+1/3^2+1/4^2+....+1/10^2 < 1
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a) Ta có: \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)
\(\Leftrightarrow2\cdot A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)
\(\Leftrightarrow2\cdot A-A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right)\)
\(\Leftrightarrow A=1-\frac{1}{2^{100}}\)
a>
\(\frac{1}{2^2}+\frac{1}{100^2}\)=1/4+1/10000
ta có 1/4<1/2(vì 2 đề bài muốn chứng minh tổng đó nhỏ 1 thì chúng ta phải xét xem có bao nhiêu lũy thừa hoặc sht thì ta sẽ lấy 1 : cho số số hạng )
1/100^2<1/2
=>A<1
Ta có : D = \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{10^2}=\frac{1}{2.2}+\frac{1}{3.3}+...+\frac{1}{10.10}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}=1-\frac{1}{10}< 1\)
=> D < 1 (đpcm)
Ta có : \(\frac{1}{2^2}< \frac{1}{1.2}\)
\(\frac{1}{3^3}< \frac{1}{2.3}\)
\(\frac{1}{4^2}< \frac{1}{3.4}\)
...
\(\frac{1}{10^2}< \frac{1}{9.10}\)
=)) \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{10^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)
Mà \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}=\frac{9}{10}< 1\)
=)) A < 1 (đpcm)
\(\Leftrightarrow2-\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{11}}\right)>0\)
Ta có: \(\frac{1}{2^{12}}-1=\left(\frac{1}{2}-1\right)\left(\frac{1}{2^{11}}+\frac{1}{2^{10}}+\frac{1}{2^9}+...+\frac{1}{2}+1\right)\)
\(\Rightarrow1+\frac{1}{2}+...+\frac{1}{2^{11}}=2\left(1-\frac{1}{2^{12}}\right)=2-\frac{1}{2^{11}}\)
\(\Rightarrow2-\left(1+\frac{1}{2}+...+\frac{1}{2^{11}}\right)=2-\left(2-\frac{1}{2^{11}}\right)=\frac{1}{2^{11}}>0\left(đpcm\right)\)
1-1/2-1/2^2-......-1/2^11
ta có:1-1/2-1/2^2-.....-1/2^11=1-(1/2+1/2^2+....+1/2^11)
A=1/2+1/2^2+1/2^3+...+1/2^11
2A=2.(1/2+1/2^2+1/2^3+...+1/2^11)
2A=2.1/2+2.1/2^2+....+2.1/2^11
2A-A=(1+1/2^2+1/2^3+...+1/2^10)-(1/2+1/2^2+1/2^3+....+1/2^11)
A=1-1/2^11=2048/2048-1/2048=2047/2048
vì 1-(1/2+1/2^2+1/2^3+...+1/2^11)=1-A
=> 1-(1/2+1/2^2+1/2^3+...+1/2^11)=1-2047/2048=2048/2048-2047/2048=1/2048=1/2^11
vậy 1-1/2-1/2^2-1/2^3-...-1/2^11=1/2^11
Ta có \(\dfrac{1}{2^2}< \dfrac{1}{1.2};\dfrac{1}{3^2}< \dfrac{1}{2.3};...;\dfrac{1}{10^2}< \dfrac{1}{9.10}\)
cộng vế với vê sta đc
\(A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}=1-\dfrac{1}{10}=\dfrac{9}{10}< 1\)
Vậy ta có đpcm
A la dat tren tong
We have: A = 1/2 ^ 2 + 1/3 ^ 2 + 1/4 ^ 2 + ........... + 1/10 ^ 2
A = 1 / 2.2 + 1 / 3.3 + 1 / 4.4 + ....... + 1 / 10:10
A <1 / 1.2 + /2.3 + 1/3.4 +......+1/9.10
A < 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ....+ 1/9 - 1/10
A < 1-1/10
Ma 1 - 1/10 = 9/10 < 1
=>A < 1 (dpcm)
dễ
1/2^2=1/1.2
1/3^2=1/2.3
1/4^2=1/3.4
....
1/10^2=1/9.10
1/1.2+1/2.3+1/3.4+...+1/9.10
=(1-1/2+1/2-1/3+1/3-1/4+...+1/9-1/10)
=1-1/10
=9/10
d=1/2^2+1/3^2+1/4^2+...+1/10^2<1/1.2+1/2.3+1/3.4+...+1/9.10
=1-1/2+1/2-1/3+1/3-1/4+...+1/9-1/10
=1-1/10<1
=>ĐPCM
nhớ **** cho mình nha