Cho \(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{99^2}+\frac{1}{100^2}\)
So sánh S với 1
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Cho
\(S=\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^{ }3}-\frac{4}{3^{ }4}+...+\frac{99}{3^{ }99}-\frac{100}{3^{ }100}\)
So sánh S và \(\frac{1}{5}\)
Cho S=\(\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{98}+\frac{1}{99}+\frac{1}{100}\)
So sánh với 1/2
Ai nhanh cho 3 tick!!!!!!!!!!!!!!!!!!!!!!!!!!!
Ta có:
\(\frac{1}{51}>\frac{1}{100}\)
\(\frac{1}{52}>\frac{1}{100}\)
...
\(\frac{1}{99}>\frac{1}{100}\)
\(\frac{1}{100}=\frac{1}{100}\)
=> S = \(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}+\frac{1}{100}>\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}+\frac{1}{100}\)
Mà số số hạng của S là: (100 - 51) : 1 + 1 = 50 (số)
=> S \(>\frac{1}{100}.50\)
=> S \(>\frac{1}{2}\)
Vậy S > 1/2.
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cho S=\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{99^2}+\frac{1}{100^2}.sosanhSvoi\frac{3}{4}\)
Cho \(S=\frac{1}{50}+\frac{1}{51}+...+\frac{1}{99}+\frac{1}{100}\) . So sánh S với \(\frac{1}{2}\)
So sánh A với 2 biết A = \(\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{99!}+\frac{1}{100!}\).
\(A=\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{99!}+\frac{1}{100!}\)
\(A< 1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\)
\(A< 1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
\(A< 1+1-\frac{1}{100}\)
\(A< 2-\frac{1}{100}< 2\)
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\(A=\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{99!}+\frac{1}{100!}\)
\(A< 1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\)
\(A< 1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
\(A< 1+1-\frac{1}{100}\)
\(A< 2-\frac{1}{100}< 2\)
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Tính và so sánh: \(S=\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}...+\frac{99}{49^2.50^2}\)\(T=\frac{1}{2^2-1^2}+\frac{1}{3^2-1^2}+\frac{1}{4^2-1^2}+...+\frac{1}{50^2-1^2}\)
\(S=\frac{3}{1^2\cdot2^2}+\frac{5}{2^2\cdot3^2}+\frac{7}{3^2\cdot4^2}+...+\frac{99}{49^2\cdot50^2}\)
\(=\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+.....+\frac{1}{49^2}-\frac{1}{50^2}\)
\(=1-\frac{1}{50^2}=\frac{2499}{2500}\)
\(T=\frac{1}{\left(2-1\right)\left(2+1\right)}+\frac{1}{\left(3-1\right)\left(3+1\right)}+...+\frac{1}{\left(50-1\right)\left(50+1\right)}\)
\(=\frac{1}{1\cdot3}+\frac{1}{2\cdot4}+\frac{1}{3\cdot5}+...+\frac{1}{49\cdot51}\)
\(=\frac{1}{2}\cdot\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{51}\right)\)
\(=\frac{1}{2}\cdot\left(1+\frac{1}{2}-\frac{1}{51}\right)=\frac{151}{204}\)
Vì \(\frac{2499}{2500}>\frac{151}{204}\)nên S>T
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JOKER_Võ Văn Quốc, T = \(\frac{1}{2}.\left(1-\frac{1}{51}+\frac{1}{2}-\frac{1}{50}\right)\)mới đúng
Sẽ dễ hơn nếu bạn chia ra 2 vế \(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\)và \(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{48+50}\)
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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 Sfrac{3}{10}+frac{3}{11}+frac{3}{12}+frac{3}{13}+frac{3}{14}
Chứng minh 1 S 2
Đọc tiếp
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 \(1< S< 2\)
Cho \(A=\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+...+\frac{99}{2^{99}}+\frac{100}{2^{100}}\)
so sánh với 2
Ta có: \(\frac{1}{2}A=\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{100}{2^{101}}\)
\(A-\frac{1}{2}A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}-\frac{100}{2^{101}}\)
Ta có: \(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{100}}=1-\frac{1}{2^{100}}< 1\)
\(\Rightarrow\frac{1}{2}A< 1-\frac{100}{2^{101}}\)
\(\Rightarrow A< 2-\frac{200}{2^{101}}< 2\)
Vậy A<2
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so sánh hai số:A=1 và B=\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+.....+\frac{1}{99^2}+\frac{1}{100^2}\)
Ta có:B=1/2^2+1/3^2+...+1/100^2<1/1*2+1/2*3+...+1/99*100
B<1-1/100<1
Mà A=1
Nên B<A
k cho mình với nha
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