B=1/3+1/32+1/33+...+1/32005
so sanh B voi 1/2
Những câu hỏi liên quan
So sanh A voi 1:
A=1/2*2 + 1/3*3 + 1/4*4 + .....+1/2011*2011
So sanh B voi 3/4:
B=1/2*2 + 1/3*3 +1/4*4 + ......+1/2011*2011
sao sanh A=\(\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\) ) voi B=\(3^{32}\)
A*2=(3-1)*(3+1)*(3^2+1)*....*(3^16+1)
A*2=(3^2-1)*(3^2+1)*(3^4+1)....*(3^16+1)
A*2=((3^4)^2-1^2)*(3^4+1)......*(3*16+1)
2*A=(3^8-1)*...(3^16+1)
bạn lm tiếp nha
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\(B=3^{32}-1=\left(3^{16}+1\right)\left(3^{16}-1\right)\)
\(=\left(3^{16}+1\right)\left(3^8+1\right)\left(3^8-1\right)\)
\(=\)\(\left(3^{16}+1\right)\left(3^8+1\right)\left(3^4+1\right)\left(3^4-1\right)\)
\(=\left(3^{16}+1\right)\left(3^8+1\right)\left(3^{\text{4}}+1\right)\left(3^2+1\right)\left(3^2-1\right)\)
\(=\left(3^{16}+1\right)\left(3^8+1\right)\left(3^4+1\right)\left(3^2+1\right)\left(3+1\right)\left(3-1\right)\)
\(=2\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(B=2A\)
đúng 100% k nha
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so sanh S=1/30+1/31+1/32+....+1/59+1/60 voi 1/2
S=1/30+1/31+1/32+1/33+...+1/59+1/60
S có 31 phân số,ta thấy:
1/30>1/62 1/31>1/62 1/32>1/62 ............ 1/60>1/62
Vậy:
S>31.1/62
S>31/62
S>1/2
Vậy S>1/2
Chúc em học tốt^^
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So sanh :
A = 1/2 + 1/4+ 1/8 + 1/16 + 1/32 voi 2005/2006
A = 31/32
Ta có 1 - 31/32 = 1/32
1 - 2005/2006 = 1/2006
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Chứng minh rằng:
A = 1/3 + 1/32 + 1/33 + ..........+ 1/399 < 1/2
B = 3/12x 22 + 5/22 x 32 + 7/32 x 42 +............+ 19/92 x 102 < 1
C = 1/3 + 2/32 + 3/33 + 4/34 +.........+ 100/3100 ≤ 0
\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{99}}\)
\(\Rightarrow\dfrac{A}{3}=\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\)
\(\Rightarrow A-\dfrac{A}{3}=\dfrac{2A}{3}=\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\dfrac{2A}{3}=\left(\dfrac{1}{3^2}-\dfrac{1}{3^2}\right)+\left(\dfrac{1}{3^3}-\dfrac{1}{3^3}\right)+...+\left(\dfrac{1}{3^{99}}-\dfrac{1}{3^{99}}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)=\dfrac{1}{3}-\dfrac{1}{3^{100}}\)
\(\Rightarrow2A=3\cdot\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\text{A}=\dfrac{1-\dfrac{1}{3^{99}}}{2}\)
\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}< \dfrac{1}{2}\)
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Khong lam phep tinh hay so sanh:
a.[-1]*[-2]*[-3]*...*[-2009] voi 0
b.[-1]*[-2]*[-3]*...*[-10] voi 1*2*3*...*9*10
cho A=(1/22-1)+(1/32-1)+...................+(1/20182-1) voi B =-1/2
so sanh A va B
cho S=1+2+2^2+.......+2^64 so sanh S voi 5.2^32
so sanh
A = 1+ 1/2 + (1/2)2+ (1/2)3+ ...+ (1/2)100 voi B = 2