\(S=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{100}}\) so sánh S với \(\dfrac{1}{2}\)
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9 tháng 5 2022 lúc 20:42
Cho S = \(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+....+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\) so sánh S và \(\dfrac{1}{5}\)
S=\(\dfrac{2}{2021+1}+\dfrac{2^2}{2021^2+1}+\dfrac{2^3}{2021^{2^2}}+...+\dfrac{2^{n+1}}{2021^{2^n}+1}+...+\dfrac{2^{2021}}{2021^{2^{2020}}+1}\)so sánh S với \(\dfrac{1}{1010}\)
\(S=\dfrac{2}{2021+1}+\dfrac{2^2}{2021^2+1}+\dfrac{2^3}{2021^{2^2}+1}+...+\dfrac{2^{n+1}}{2021^{2^n}+1}+...+\dfrac{2^{2021}}{2021^{2^{2020}}+1}\)
So sánh S với \(\dfrac{1}{1010}\)
1. Chứng minh:
\(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}< \dfrac{3}{16}\)
2. Cho:
\(M=\dfrac{1}{15}+\dfrac{1}{105}+\dfrac{1}{315}+...+\dfrac{1}{1977}\). So sánh M với 12.
Bài 2: A = \(\dfrac{x^2+2}{x^3-1}+\dfrac{x+1}{x^2+x+1}\) và B = \(\dfrac{1}{x-1}\)
a) Tính giá trị của B khi \(x^2-8x+7=0\)
b) Chứng tỏ A = \(\dfrac{2x^2+1}{x^3-1}\)
c) Rút gọn S = A - B
d) Tìm x để S = \(\dfrac{1}{3}\)
e) So sánh S với $\frac{1}{3}$
a) ĐKXĐ: \(x\ne1\)
Ta có: \(x^2-8x+7=0\)
\(\Leftrightarrow x^2-x-7x+7=0\)
\(\Leftrightarrow x\left(x-1\right)-7\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(loại\right)\\x=7\left(nhận\right)\end{matrix}\right.\)
Thay x=7 vào B, ta được:
\(B=\dfrac{1}{7-1}=\dfrac{1}{6}\)
Vậy: Khi \(x^2-8x+7=0\) thì \(B=\dfrac{1}{6}\)
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b) Ta có: \(A=\dfrac{x^2+2}{x^3-1}+\dfrac{x+1}{x^2+x+1}\)
\(=\dfrac{x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{\left(x+1\right)\left(x-1\right)}{\left(x^2+x+1\right)\left(x-1\right)}\)
\(=\dfrac{x^2+2+x^2-1}{x^3-1}\)
\(=\dfrac{2x^2+1}{x^3-1}\)
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c) Ta có: S=A-B
\(=\dfrac{2x^2+1}{x^3-1}-\dfrac{1}{x-1}\)
\(=\dfrac{2x^2+1}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{2x^2+1-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{x^2-x}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{x}{x^2+x+1}\)
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So sánh \(A\) với \(\dfrac{3}{4}\), biết \(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}\)
1/32< 1/2.3
1/42< 1/3.4
...
1/1002< 1/99.100
=> 1/22 + 1/32 + 1/42 + ... + 1/1002< 1/22 + 1/2.3 + 1/3.4 + ... + 1/99.100
A < 1/4 + 1/2 -1/3 + 1/3 - 1/4 +... + 1/99 - 1/100
A < 1/4 + 1/2 -1/100 < 1/4 + 1/2 = 3/4
=> A < 3/4
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So sánh tổng S= \(\dfrac{1}{2}\)+\(\dfrac{2}{2^2}\)+\(\dfrac{3}{2^3}\)+...+\(\dfrac{n}{2^n}\)+...+\(\dfrac{2017}{2^{2017}}\)với 2 (\(n\in N\)*)
Giải:
\(S=\dfrac{1}{2}+\dfrac{2}{2^2}+...+\dfrac{n}{2^n}+...+\dfrac{2017}{2^{2017}}\)
Với \(n>2\) thì \(\dfrac{n}{2^n}=\dfrac{n+1}{2^{n-1}}-\dfrac{n+2}{2^n}\)
Ta có:
\(\dfrac{n+1}{2^{n-1}}=\dfrac{n+1}{2^n:2}=\dfrac{2.\left(n+1\right)}{2^n}\)
\(\Rightarrow\dfrac{n+1}{2^{n-1}}-\dfrac{n+2}{2^n}\)
\(=\dfrac{2.\left(n+1\right)}{2^n}-\dfrac{n+2}{2^n}\)
\(=\dfrac{2.\left(n+1\right)-n-2}{2^n}\)
\(=\dfrac{n}{2^n}\)
\(\Leftrightarrow S=\dfrac{1}{2}+\left(\dfrac{2+1}{2^{2-1}}-\dfrac{2+2}{2^2}\right)+...+\left(\dfrac{2016+1}{2^{2015}}-\dfrac{2018}{2^{2016}}\right)+\left(\dfrac{2017+1}{2^{2016}}-\dfrac{2019}{2^{2017}}\right)\)
\(S=\dfrac{1}{2}+\dfrac{3}{2}+\dfrac{2019}{2017}\)
\(S=2-\dfrac{2019}{2017}\)
\(\Leftrightarrow S=2-\dfrac{2019}{2017}< 2\)
Hay \(S< 2\)
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So sánh
\(A=\dfrac{1}{2^2}+\dfrac{2}{2^3}+\dfrac{3}{2^4}+...+\dfrac{100}{2^{201}}\)
và \(B=\dfrac{1}{2^3}+\dfrac{1}{3^3}+\dfrac{1}{4^3}+...+\dfrac{1}{100^3}\)
28 tháng 6 2021 lúc 15:47
So sánh A và B :
\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{100}}\)
\(B=\dfrac{1}{2}\)
Ta có `3A=1+1/3+....+1/3^99`
`=>3A-A=1-1/3^100`
`=>2A=1-1/3^100`
`=>A=1/2-1/(2.3^100)<1/2`
Hay `A<B`
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