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2 vế bằng nhau
100-(1+1/2+1/3+...+1/100) = 1/2+2/3+3/4+...+99/100
100- 1-1/2-1/3-...-1/100 = 1/2+2/3+3/4+...+99/100
100 = 1 + 1/2 + 1/2 + 1/3 + 2/3 + ... + 1/100 + 99/100 (cùng cộng 2 vế với (- 1-1/2-1/3-...-1/100)
100 = 1 + 1 + 1 + ... + 1 (100 số hạng)
100 = 100
Vậy 100-(1+1/2+1/3+...+1/100) = 1/2+2/3+3/4+...+99/100
Ta có
\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)
..............
\(\dfrac{1}{100^2}< \dfrac{1}{99.100}\)
=> S < \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)
S < \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(S< 1-\dfrac{1}{100}< 1\)(do 1/100 >0)
ĐPcm
Giải:
\(S=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{99^2}+\dfrac{1}{100^2}\)
Ta có:
\(\dfrac{1}{2^2}=\dfrac{1}{2.2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}=\dfrac{1}{3.3}< \dfrac{1}{2.3}\)
\(\dfrac{1}{4^2}=\dfrac{1}{4.4}< \dfrac{1}{3.4}\)
\(...\)
\(\dfrac{1}{99^2}=\dfrac{1}{99.99}< \dfrac{1}{98.99}\)
\(\dfrac{1}{100^2}=\dfrac{1}{100.100}< \dfrac{1}{99.100}\)
\(\Rightarrow S< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{98.99}+\dfrac{1}{99.100}\)
\(\Rightarrow S< \dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\)
\(\Rightarrow S< \dfrac{1}{1}-\dfrac{1}{100}< 1\)
\(\Rightarrow S< 1\)
Vậy S < 1.
\(P=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}\)
\(3P=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}\)
\(3P-P=\left(1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}\right)-\left(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}\right)\)\(2P=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
\(6P=3+1+\frac{1}{3}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\)
\(6P-2P=\left(3+1+\frac{1}{3}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\right)\)
\(4P=3-\frac{100}{3^{99}}-\frac{1}{3^{99}}+\frac{100}{3^{100}}\)
\(4P=3-\frac{300}{3^{100}}-\frac{3}{3^{100}}+\frac{100}{3^{100}}\)
\(4P=3-\frac{303}{3^{100}}+\frac{100}{3^{100}}\)
\(4P=3-\frac{203}{3^{100}}< 3\)
\(P< \frac{3}{4}\)
\(P< Q\)
\(\frac{4}{3}.\left(\frac{1}{6}-\frac{1}{2}\right)\le x\le\frac{2}{3}.\left(\frac{-1}{6}+\frac{3}{4}\right)\)
\(\frac{4}{3}.\frac{-1}{3}\le x\le\frac{2}{3}.\frac{7}{12}\)
\(\frac{-4}{9}\le x\le\frac{7}{18}\)
\(\frac{-8}{18}\le x\le\frac{7}{18}\)
\(\Rightarrow\)X \(\in\) {\(\frac{-7}{18};\frac{-6}{18};\frac{-5}{18};\frac{-4}{18};\frac{-3}{18};\frac{-2}{18};\frac{-1}{18};0;\frac{1}{18};\frac{2}{18};\frac{3}{18};\frac{4}{18};\frac{5}{18};\frac{6}{18}\)}
giúp toán mik trc đi mk giải cho