Cho \(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}};B=\frac{1}{2}\).so sánh A và B
Chứng minh :
a) \(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{3}{16}\) \(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{4^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{3}{16}\)
b)\(\frac{1}{41}+\frac{1}{42}+\frac{1}{43}+...+\frac{1}{79}+\frac{1}{80}< \frac{7}{12}\)
c) Cho \(S=\frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}\)
Chứng minh \(1< S< 2\)
Chứng minh rằng:
a. \(\frac{1}{3^2}+\frac{2}{3^3}+\frac{3}{3^4}+\frac{4}{3^5}+...+\frac{99}{3^{100}}+\frac{100}{3^{101}}< \frac{1}{4}\)
b.\(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}< \frac{1}{3}\)
c.\(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{1}{16}\)
d. \(\frac{1}{5^2}-\frac{2}{5^3}+\frac{3}{5^4}-\frac{4}{5^5}+...+\frac{99}{5^{100}}-\frac{100}{5^{101}}< \frac{1}{36}\)
Cho biểu thức: A=\(-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+\frac{1}{3^4}-\frac{1}{3^5}+...+\frac{1}{3^{100}}\)
Tính giá trị biểu thức B=\(4|A|+\frac{1}{3^{100}}\)
Cho biểu thức: \(A=-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+\frac{1}{3^4}-\frac{1}{3^5}+...+\frac{1}{3^{100}}\)
Tính giá trị biểu thức \(B=4|A|+\frac{1}{3^{100}}\)
\(A=\left(\frac{1}{3^2}+\frac{1}{3^4}+...+\frac{1}{3^{100}}\right)-\left(\frac{1}{3}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\right)\)
\(3^2A=3^2\left(\frac{1}{3^2}+\frac{1}{3^4}+...+\frac{1}{3^{100}}\right)-3^2\left(\frac{1}{3}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\right)\)
\(9A=\left(1+\frac{1}{3^2}+...+\frac{1}{3^{98}}\right)-\left(3+\frac{1}{3}+...+\frac{1}{3^{97}}\right)\)
\(9A-A=\left(1-\frac{1}{3^{100}}\right)-\left(3-\frac{1}{3^{99}}\right)\)
\(8A=1-3=-2\)
A=\(\frac{-2}{8}=\frac{-1}{4}\)
\(B=4\left|\frac{-1}{4}\right|+\frac{1}{3^{100}}=1+\frac{1}{3^{100}}=1\)
Vậy B=1
Trl:
Bạn kia trả lời đúng rồi nhoa : ))
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Bạn kia trả lời sai rồi
Cho;A=\(\frac{1}{3^2}-\frac{1}{3^4}+\frac{1}{3^6}-\frac{1}{3^8}+......+\frac{1}{3^{98}}-\frac{1}{3^{100}}\)
Ta có : \(A=\frac{1}{3^2}-\frac{1}{3^4}+\frac{1}{3^6}-\frac{1}{3^8}+....+\frac{1}{3^{98}}-\frac{1}{3^{100}}\)(1)
=> 32.A = \(1-\frac{1}{3^2}+\frac{1}{3^4}-\frac{1}{3^8}+...+\frac{1}{3^{96}}-\frac{1}{3^{98}}\)(2)
Lấy (2) cộng (1) theo vế ta có :
32.A + A = \(\left(\frac{1}{3^2}-\frac{1}{3^4}+\frac{1}{3^6}-\frac{1}{3^8}+...+\frac{1}{3^{98}}-\frac{1}{3^{100}}\right)+\left(1-\frac{1}{3^2}+\frac{1}{3^4}-\frac{1}{3^6}+...+\frac{1}{3^{96}}-\frac{1}{3^{98}}\right)\)
10A = \(1-\frac{1}{3^{100}}\)
=> A = \(\left(1-\frac{1}{3^{100}}\right):10=\frac{1}{10}-\frac{1}{3^{100}.10}=0,1-\frac{1}{3^{100}.10}< 0,1\)
=> A < 0,1 (ĐPCM)
\(A=\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)
\(\Rightarrow3A=1-\frac{2}{3}+\frac{3}{3^2}-...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\)
\(\Rightarrow3A+A=\left(...\right)+\left(...\right)\)
\(\Rightarrow4A=1-\frac{1}{3}+\frac{1}{3^2}-...-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
\(\Rightarrow3.4A=3-1+\frac{1}{3}-...-\frac{1}{3^{98}}-\frac{100}{3^{99}}\)
\(\Rightarrow12A+4A=\left(...\right)+\left(...\right)\)
\(\Rightarrow16A=3-\frac{101}{3^{99}}-\frac{100}{3^{100}}< 3\)
\(\Rightarrow A< \frac{3}{16}\)
1) \(+2x+3y⋮17\)
\(\Rightarrow26x+39y⋮17\)
\(\Rightarrow\left(9x+5y\right)+17x+34y⋮17\)
Mà \(17x+34y⋮17\)
\(\Rightarrow9x+5y⋮17\)
\(+9x+5y⋮17\)
\(\Rightarrow36x+20y⋮17\)
\(\Rightarrow\left(2x+3y\right)+34x+17y⋮17\)
Mà \(34x+17y⋮17\)
\(\Rightarrow2x+3y⋮17\)
Mong mọi người giúp em với ạ!
a) Cho A=\(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+.....+\frac{100}{3^{100}}\)Chứng minh A<\(\frac{3}{4}\).
b) Chứng minh rằng:A=\(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+.....+\frac{1}{3^{99}}< \frac{1}{2}\)
b) A=\(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\)
3A=\(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)
3A-A=\(1-\frac{1}{3^{99}}\)
2A=\(1-\frac{1}{3^{99}}\)
vì 2A<1
=> A<\(\frac{1}{2}\)
cho biểu thức:
A= \(-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+\frac{1}{3^4}-\frac{1}{3^5}+...+\frac{1}{3^{100}}\)
Tính giá trị của biểu thức B=4|A|+\(\frac{1}{3^{100}}\)
cho biểu thức \(A=-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+\frac{1}{3^4}-\frac{1}{3^5}+...+\frac{1}{3^{100}}\)
tính giá trị biểu thức \(B=4\left|A\right|+\frac{1}{3^{100}}\)
Chứng tỏ rằng
a) \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< 1\)
b) \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2018^2}< \frac{3}{4}\)
c) \(\frac{1}{3}+\frac{3}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}< \frac{3}{4}\)
\(a)\) Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\) ta có :
\(A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(A< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(A< 1-\frac{1}{100}=\frac{99}{100}< 1\)
Vậy \(A< 1\)
Chúc bạn học tốt ~