Cm
S= 1/22-1/24+1/26-...+1/22002-1/22004<0.2
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Ta có:
\(A=1+2+2^2+...+2^{2002}\)
\(2A=2+2^2+2^3+...+2^{2003}\)
\(2A-A=\left(2+2^2+2^3+...+2^{2003}\right)-\left(1+2+2^2+....+2^{2002}\right)\)
\(A=2^{2003}-1\)
Mà: \(2^{2003}=2^{2003}\)
\(\Rightarrow2^{2003}-1< 2^{2003}\)
\(\Rightarrow A< B\)
A = 2 + 22 + 23 + … + 22004 . Chứng minh rằng A chia hết cho 3 , cho 7.
Đặt :
\(A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+.....+\dfrac{1}{2^{99}}\)
\(\Leftrightarrow2A=3+\dfrac{1}{2}+\dfrac{1}{2^2}+....+\dfrac{1}{2^{98}}\)
\(\Leftrightarrow2A-A=\left(3+\dfrac{1}{2}+....+\dfrac{1}{2^{98}}\right)-\left(1+\dfrac{1}{2}+....+\dfrac{1}{2^{99}}\right)\)
\(\Leftrightarrow A=2-\dfrac{1}{2^{99}}\)
Vậy..
Số số hạng của tổng A là : \(\dfrac{30-21}{1}+1=10\left(sh\right)\)
`=>A=\underbrace{1/21+1/22+...+1/30}_{10sh}>\underbrace{1/30+1/30+1/30+...+1/30}_{10sh}`
`=>A>(1)/(30).10`
`=>A>10/30`
`=>A>1/3`
`=>đpcm`
\(S=1+2^2+2^4+2^6+...+2^{100}\)
\(2^2S=2^2\left(1+2^2+2^4+2^6+...+2^{100}\right)\)
\(4S=2^2+2^4+2^6+2^8+...+2^{102}\)
\(4S-S=\left(2^2+2^4+2^6+2^8+...+2^{102}\right)-\left(1+2^2+2^4+2^6+...+2^{100}\right)\)
\(3S=2^{102}-1\)
\(S=\dfrac{2^{102}-1}{3}\)
Ta có: \(\dfrac{x^{24}+x^{20}+x^{16}+...+x^4+1}{x^{26}+x^{24}+x^{22}+...+x^2+1}\)
\(=\dfrac{x^{24}+x^{20}+x^{16}+...+x^4+1}{\left(x^{26}+x^{22}+...+x^2\right)+\left(x^{24}+x^{20}+x^{16}+...+x^4+1\right)}\)
\(=\dfrac{x^{24}+x^{20}+x^{16}+...+x^4+1}{x^2\left(x^{24}+x^{20}+...+1\right)+\left(x^{24}+x^{20}+x^{16}+...+x^4+1\right)}\)
\(=\dfrac{x^{24}+x^{20}+x^{16}+...+x^4+1}{\left(x^{24}+x^{20}+x^{16}+...+1\right)\left(x^2+1\right)}\)
\(=\dfrac{1}{x^2+1}\)
x24+x20+x16+...+x4+1x26+x24+x22+...+x2+1x24+x20+x16+...+x4+1x26+x24+x22+...+x2+1
=x24+x20+x16+...+x4+1(x26+x22+...+x2)+(x24+x20+x16+...+x4+1)=x24+x20+x16+...+x4+1(x26+x22+...+x2)+(x24+x20+x16+...+x4+1)
=x24+x20+x16+...+x4+1x2(x24+x20+...+1)+(x24+x20+x16+...+x4+1)=x24+x20+x16+...+x4+1x2(x24+x20+...+1)+(x24+x20+x16+...+x4+1)
=x24+x20+x16+...+x4+1(x24+x20+x16+...+1)(x2+1)
A=4+22+23+....+220
2A=8+23+24+...+221
=> A+2A-A = (8+23+24+...+221) - (4+22+23+....+220)
=>A=221+8 - (22+4)=221
=>A là 1 lũy thừa của 2
1/
Tổng A là tổng các số hạng cách đều nhau 4 đơn vị.
Số số hạng: $(101-1):4+1=26$
$A=(101+1)\times 26:2=1326$
2/
$B=(1+2+2^2)+(2^3+2^4+2^5)+(2^6+2^7+2^8)+(2^9+2^{10}+2^{11})$
$=(1+2+2^2)+2^3(1+2+2^2)+2^6(1+2+2^2)+2^9(1+2+2^2)$
$=(1+2+2^2)(1+2^3+2^6+2^9)$
$=7(1+2^3+2^6+2^9)\vdots 7$