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\(S=\dfrac{1}{2^2}\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}\right)\)
=>\(S< =\dfrac{1}{4}\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\right)\)
=>\(S< =\dfrac{1}{4}\cdot\left(1-\dfrac{1}{n}\right)=\dfrac{1}{4}\cdot\dfrac{n-1}{n}< =\dfrac{1}{4}\)
Từ đề có:
\(\dfrac{2-1}{2!}\) + \(\dfrac{3-1}{3!}\) + .... + \(\dfrac{2014-1}{2014!}\)
= \(\dfrac{2}{2!}\) - \(\dfrac{1}{2!}\) + \(\dfrac{3}{3!}\) - \(\dfrac{1}{3!}\) + .... + \(\dfrac{2014}{2014!}\) - \(\dfrac{1}{2014!}\)
= 1 - \(\dfrac{1}{2!}\) + \(\dfrac{1}{2!}\) - \(\dfrac{1}{3!}\) + .... + \(\dfrac{1}{2013!}\) - \(\dfrac{1}{2014!}\)
= 1 - \(\dfrac{1}{2014!}\), rứa đủ rồi đúng không ?
Có chi không hiểu mai ta giảng cho nhớ tick đúng nha
a) Vế trái \(=\dfrac{1.3.5...39}{21.22.23...40}=\dfrac{1.3.5.7...21.23...39}{21.22.23....40}=\dfrac{1.3.5.7...19}{22.24.26...40}\)
\(=\dfrac{1.3.5.7....19}{2.11.2.12.2.13.2.14.2.15.2.16.2.17.2.18.2.19.2.20}\\ =\dfrac{1.3.5.7.9.....19}{\left(1.3.5.7.9...19\right).2^{20}}=\dfrac{1}{2^{20}}\left(đpcm\right)\)
b) Vế trái
\(=\dfrac{1.3.5...\left(2n-1\right)}{\left(n+1\right).\left(n+2\right).\left(n+3\right)...2n}\\ =\dfrac{1.2.3.4.5.6...\left(2n-1\right).2n}{2.4.6...2n.\left(n+1\right)\left(n+2\right)...2n}\\ =\dfrac{1.2.3.4...\left(2n-1\right).2n}{2^n.1.2.3.4...n.\left(n+1\right)\left(n+2\right)...2n}\\ =\dfrac{1}{2^n}.\\ \left(đpcm\right)\)
a, Gọi phân số cần tìm là \(\dfrac{a}{b}\); phân số sau khi cộng là \(\dfrac{a+b}{b}\).
Theo bài ra ta có ;
\(\dfrac{a}{b}\cdot7=\dfrac{a+b}{b}\\ \Leftrightarrow\dfrac{7a}{b}=\dfrac{a}{b}+1\\ \Leftrightarrow\dfrac{7a}{b}-\dfrac{a}{b}=1\\ \Leftrightarrow\dfrac{6a}{b}=1\\ \Leftrightarrow6a=b\)
Vì \(\dfrac{a}{b}\) là phân số tối giản nên \(\dfrac{a}{b}=\dfrac{1}{6}\)
Vậy phân số tối giản cần tìm là \(\dfrac{1}{6}\)
b, Đặt \(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{n^2}\)
Ta có :
\(A< \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{\left(n-1\right)\cdot n}=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\\ =\dfrac{1}{2}-\dfrac{1}{n}\)
Vì \(n\ge2vàn\in N\Rightarrow\dfrac{1}{2}\ge\dfrac{1}{n}\Rightarrow\dfrac{1}{2}-\dfrac{1}{n}< \dfrac{1}{2}\)
Mà \(\dfrac{1}{2}< \dfrac{97}{144}\Rightarrow\dfrac{1}{2}-\dfrac{1}{n}< \dfrac{97}{144}\Leftrightarrow A< \dfrac{97}{144}\\ \RightarrowĐpcm\)
\(Vì\dfrac{1}{2^2}< \dfrac{1}{1.2};\dfrac{1}{3^2}< \dfrac{1}{2.3};.....;\dfrac{1}{n^2}< \dfrac{1}{(n-1).n}\)
\(\Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+....+\dfrac{1}{n^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+....+\dfrac{1}{\left(n-1\right).1}< 1\)\(\Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+....+\dfrac{1}{n^2}< 1\left(đpcm\right)\)
vậy \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+....+\dfrac{1}{n^2}< 1\)
ĐặtA= \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+....+\dfrac{1}{n^2}\)
Do \(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)
.............
\(\dfrac{1}{n^2}< \dfrac{1}{\left(n-1\right)n}\)
Cộng vế với vế ta suy ra : A<\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+.......+\dfrac{1}{\left(n-1\right)n}=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-....-\dfrac{1}{\left(n-1\right)}+\dfrac{1}{n-1}-\dfrac{1}{n}\)
=\(1-\dfrac{1}{n}\)
Mà 1-\(\dfrac{1}{n}\)<1
=> A<1 (đpcm)
Đặt :
\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+..........+\dfrac{1}{n^2}\)
Ta thấy :
\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)
..........................
\(\dfrac{1}{n^2}< \dfrac{1}{\left(n-1\right)n}\)
\(\Leftrightarrow A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+..........+\dfrac{1}{\left(n-1\right)n}\)
\(\Leftrightarrow A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...........+\dfrac{1}{n-1}-\dfrac{1}{n}\)
\(\Leftrightarrow A< 1-\dfrac{1}{n}< 1\)
\(\Leftrightarrow A< 1\)
Vậy ......
Gọi tổng trên là A
1/2.2<1/1.2
1/3.3<1/2.3
........
1/n.n<1/(n-1).n
=>A< 1/1.2+1/2.3+.....+1/(n-1).n
=> A<1-1/2+1/2-1/3+....+1/(n-1)-1/n
=> A< 1-1/n<1
=>A<1
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