(3+1)(32+1)(34+1)(38+1)(316+1)(332+1)
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b: A=1/3+1/9+...+1/3^10
=>3A=1+1/3+...+1/3^9
=>A*2=1-1/3^10=(3^10-1)/3^10
=>A=(3^10-1)/(2*3^10)
c: C=3/2+3/8+3/32+3/128+3/512
=>4C=6+3/2+...+3/128
=>3C=6-3/512
=>C=1023/512
d: A=1/2+...+1/256
=>2A=1+1/2+...+1/128
=>A=1-1/256=255/256
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\(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(2A=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(2A=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(2A=\left(3^{16}-1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(2A=\left(3^{32}-1\right)\left(3^{32}+1\right)\)
\(2A=3^{64}-1\Rightarrow A=\dfrac{3^{64}-1}{2}\)
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Áp dụng đẳng thức : a^2 - 1 = (a + 1)(a - 1)
=> a + 1 = (a^2 - 1)/(a + 1)
Ta có: 3 + 1 = (3^2 - 1)/(3 - 1)
3^2 + 1 = (3^4 - 1)/(3^2 - 1)
3^4 + 1 = (3^8 - 1)/(3^4 - 1)
3^8 + 1 = (3^16 - 1)/(3^8 - 1)
3^16 + 1 = (3^32 - 1)/(3^16 - 1)
3^32 + 1 = (3^64 - 1)/(3^32 - 1)
(3 + 1)(3^2 + 1)(3^4 + 1)(3^8 + 1)(3^16 + 1)(3^32 + 1)
=(3^2 - 1)/(3 - 1).(3^4 - 1)/(3^2 - 1).(3^8 - 1)/(3^4 - 1).(3^32 - 1)/(3^16 - 1).(3^64 - 1)/(3^32 - 1)
=(3^64 - 1)/(3 - 1)
=(3^64 - 1)/2
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
8A=(3^2-1)(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)-4.3^{32}$
$=[(3^2-1)(3^2+1)](3^4+1)(3^8+1)(3^{16}+1)-4.3^{32}$
$=(3^4-1)(3^4+1)(3^8+1)(3^{16}+1)-4.3^{32}$
$=(3^8-1)(3^8+1)(3^{16}+1)-4.3^{32}$
$=(3^{16}-1)(3^{16}+1)-4.3^{32}$
$=3^{32}-1-4.3^{32}$
$=-3.3^{32}-1=-3^{33}-1$
$\Rightarrow A=\frac{-3^{33}-1}{8}$
![](https://rs.olm.vn/images/avt/0.png?1311)
Chứng minh rằng:
a) 1/2-1/4+1/8-1/16+1/32-1/64<1/3
b) 1/3-2/3^2+3/3^3-3/3^4+...+99/3^99-100/3^100<3/16
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