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Đặt: \(A=\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+...+\dfrac{1}{100^2}\)
Ta có: \(\left\{{}\begin{matrix}A>\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+...+\dfrac{1}{100.101}=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{101}=\dfrac{1}{5}-\dfrac{1}{101}>\dfrac{1}{6}\\A< \dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{99.100}=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{100}=\dfrac{1}{4}-\dfrac{1}{100}< \dfrac{1}{4}\end{matrix}\right.\)
Vậy \(\dfrac{1}{6}< A< \dfrac{1}{4}\)
\(A=\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}\\ =\dfrac{1}{5.5}+\dfrac{1}{6.6}+...+\dfrac{1}{100.100}\\ < \dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}\\ =\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ =\dfrac{1}{4}-\dfrac{1}{100}< \dfrac{1}{4}\)
\(A=\dfrac{1}{5^2}+\dfrac{1}{6^2}+....+\dfrac{1}{100^2}>\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{100.101}\\=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{101}\\ =\dfrac{1}{5}-\dfrac{1}{101}\)
1/6<1/5^2+1/6^2+1/7^2+...+1/100^2<1/4
ta có:
(+)1/5^2+1/6^2+1/7^2+...+1/100^2<1/4.5+1/5.6+...+1/99.100
=1/4-1/5+1/5-...+1/99-1/100
=1/4-1/100<1/4
=>1/5^2+1/6^2+1/7^2+...+1/100^2<1/4
(+)1/5^2+1/6^2+1/7^2+...+1/100^2>1/5.6+...+1/99.100
=1/5-1/6+1/6-...+1/99-1/100
=1/5-1/100>1/6
=>1/5^2+1/6^2+1/7^2+...+1/100^2
\(\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+....+\dfrac{1}{100^2}\\ >\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{100.101}\\ =\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{101}\\ =\dfrac{1}{5}-\dfrac{1}{101}\\ =\dfrac{96}{505}\\ >\dfrac{1}{6}\)
\(\dfrac{1}{5^2}+...+\dfrac{1}{100^2}\\ < \dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+....+\dfrac{1}{99.100}\\ =\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ =\dfrac{1}{4}-\dfrac{1}{100}< \dfrac{1}{4}\)
Bài 1:
a) Ta có: \(\dfrac{7^4\cdot3-7^3}{7^4\cdot6-7^3\cdot2}\)
\(=\dfrac{7^3\cdot\left(7\cdot3-1\right)}{7^3\cdot2\left(7\cdot3-1\right)}\)
\(=\dfrac{1}{2}\)
c) Ta có: \(E=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}\)
\(\Leftrightarrow\dfrac{1}{3}\cdot E=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{101}}\)
\(\Leftrightarrow E-\dfrac{1}{3}\cdot E=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{101}}\right)\)
\(\Leftrightarrow E\cdot\dfrac{2}{3}=1-\dfrac{1}{3^{101}}\)
\(\Leftrightarrow E=\dfrac{3-\dfrac{3}{3^{101}}}{2}=\dfrac{1-\dfrac{1}{3^{100}}}{2}\)
A=\(\dfrac{7^2-1}{7^4}+\dfrac{7^2-1}{7^8}+...+\dfrac{7^2-1}{7^{100}}=\left(7^2-1\right)\left(\dfrac{1}{7^4}+\dfrac{1}{7^8}+...+\dfrac{1}{7^{100}}\right)=48\cdot B\)Dễ dàng tính được B( nhân hết với 7 mũ 4 roi trừ đi, chia ra là xong) ra đpcm.
Lên lớp 11 thì ta có dạng tổng quát luôn này(tức là nếu n quá lớn thì có thể coi là xảy ra dấu bằng) \(\dfrac{1}{7^2}-\dfrac{1}{7^4}+...+\dfrac{1}{7^n}-\dfrac{1}{7^{n+2}}< \dfrac{1}{50}\)
Ta có: \(\dfrac{2}{5^2}+\dfrac{2}{6^2}+\dfrac{2}{7^2}+...+\dfrac{1}{100^2}< \dfrac{2}{4.5}+\dfrac{2}{5.6}+\dfrac{2}{6.7}+...+\dfrac{2}{99.100}\)
\(=2.\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{99.100}\right)\)
\(=2.\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(=2.\left(\dfrac{1}{4}-\dfrac{1}{100}\right)< 2.\dfrac{1}{4}=\dfrac{1}{2}\)
hay \(\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+...+\dfrac{1}{100^2}< \dfrac{1}{2}\)
Vậy ...
Gọi \(A=\dfrac{2}{5^2}+\dfrac{2}{6^2}+\dfrac{2}{7^2}+...+\dfrac{2}{100^2}\)
\(A< \dfrac{2}{4\cdot5}+\dfrac{2}{5\cdot6}+\dfrac{2}{6\cdot7}+...+\dfrac{2}{99\cdot100}\\ A< 2\cdot\left(\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+...+\dfrac{1}{99\cdot100}\right)\\ A< 2\cdot\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\\ A< 2\cdot\left(\dfrac{1}{4}-\dfrac{1}{100}\right)\\ A< \dfrac{1}{2}-\dfrac{1}{50}\\ A< \dfrac{12}{25}< \dfrac{1}{2}\)
Vậy \(\dfrac{2}{5^2}+\dfrac{2}{6^2}+\dfrac{2}{7^2}+...+\dfrac{2}{100^2}< \dfrac{1}{2}\)