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30 tháng 8 2020
a) Ta có: \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)
\(\Leftrightarrow2\cdot A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)
\(\Leftrightarrow2\cdot A-A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right)\)
\(\Leftrightarrow A=1-\frac{1}{2^{100}}\)
HV
1
20 tháng 8 2016
1/42 + 1/62 + 1/82 + ... + 1/(2n)2
= 1/22.(1/22 + 1/32 + 1/42 + ... + 1/n2)
< 1/22.(1/1.2 + 1/2.3 + 1/3.4 + ... + 1/(n-1).n)
< 1/4.(1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/n-1 - 1/n)
< 1/4.(1 - 1/n)
< 1/4.1 = 1/4 ( đpcm)
DV
1
21 tháng 6 2017
\(VT=\left(a-1\right)\left(a-2\right)\left(1+a+a^2\right)\left(4+2a+a^2\right)\)
\(=\left(a^3-1\right)\left(a^3-8\right)\)
\(=a^6-8a^3-a^3+8\)
\(=a^6-9a^3+8=VP\)
\(\Rightarrowđpcm\)
NT
1
30 tháng 7 2018
\(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=2^{32}-1\)
NM
2
AT
23 tháng 6 2017
Ta có:
\(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2+1\right)\left(2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\)
\(=2^{32}-1\)
Vậy...
Biến đổi vế trái ta có:
\(\left(2^1+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^6+1\right)\left(2^8+1\right)\)
= \(\left(2-1\right)\left(2+1\right)\left(2^4+1\right)\left(2^6+1\right)\left(2^8+1\right)\)
= \(\left(2^4-1\right)\left(2^4+1\right)\left(2^6+1\right)\left(2^8+1\right)\)
= \(\left(2^8-1\right)\left(2^6+1\right)\left(2^8+1\right)\)
= \(\left(2^{16}-1\right)\left(2^6+1\right)\)
=> Sai đề
\(VT=\left(2^1+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^6+1\right)\left(2^8+1\right)\)
\(=\left(2^1-1\right)\left(2^1+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^6+1\right)\left(2^8+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^6+1\right)\left(2^8+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^6+1\right)\left(2^8+1\right)\)
\(=\left(2^{16}-1\right)\left(2^6+1\right)\left(2^8+1\right)\)
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