chứng tỏ 1/2^2+1/3^2+1/4^2+.....+1/100^2 < 1/2
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TT
1
3 tháng 3
\(\dfrac{1}{2^2}>\dfrac{1}{2\cdot3}=\dfrac{1}{2}-\dfrac{1}{3}\)
\(\dfrac{1}{3^2}>\dfrac{1}{3\cdot4}=\dfrac{1}{3}-\dfrac{1}{4}\)
...
\(\dfrac{1}{9^2}>\dfrac{1}{9\cdot10}=\dfrac{1}{9}-\dfrac{1}{10}\)
Do đó: \(A>\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}=\dfrac{1}{2}-\dfrac{1}{10}=\dfrac{4}{10}=\dfrac{2}{5}\)
\(\dfrac{1}{2^2}< \dfrac{1}{1\cdot2}=1-\dfrac{1}{2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2\cdot3}=\dfrac{1}{2}-\dfrac{1}{3}\)
...
\(\dfrac{1}{9^2}< \dfrac{1}{8\cdot9}=\dfrac{1}{8}-\dfrac{1}{9}\)
Do đó: \(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{9^2}< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{8}-\dfrac{1}{9}=\dfrac{8}{9}\)
Suy ra: \(\dfrac{2}{5}< A< \dfrac{8}{9}\)
3 tháng 3
\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\)
=>\(3A=1+\dfrac{1}{3}+...+\dfrac{1}{3^{98}}\)
=>\(3A-A=1+\dfrac{1}{3}+...+\dfrac{1}{3^{98}}-\dfrac{1}{3}-\dfrac{1}{3^2}-...-\dfrac{1}{3^{99}}\)
=>\(2A=1-\dfrac{1}{3^{99}}\)
=>\(A=\dfrac{1}{2}-\dfrac{1}{2\cdot3^{99}}< \dfrac{1}{2}\)
3 tháng 3
\(S=\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2020}}\)
=>\(2S=1+\dfrac{1}{2}+...+\dfrac{1}{2^{2019}}\)
=>\(2S-S=1+\dfrac{1}{2}+...+\dfrac{1}{2^{2019}}-\dfrac{1}{2}-\dfrac{1}{2^2}-...-\dfrac{1}{2^{2020}}\)
=>\(S=1-\dfrac{1}{2^{2020}}< 1\)
NT
0
AN
0
3 tháng 3
\(S=\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{1001\cdot1003}\)
\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{1001}-\dfrac{1}{1003}\)
\(=1-\dfrac{1}{1003}< 1\)
3 tháng 3
a: Những tia trên hình vẽ là Ex,Ey,Em,En,Ct,CK,Cn
Đoạn thẳng: EK,EC,CK
b: Các cặp tia đối nhau là:
Ex;Ey
Kx;Ky
Cn;CE
CK,Ct
CT
1
H9
HT.Phong (9A5)
CTVHS
3 tháng 3
\(A=\dfrac{1}{2}+\dfrac{3}{2}+\left(\dfrac{3}{2}\right)^2+...+\left(\dfrac{3}{2}\right)^{2023}\)
Đặt: \(C=\dfrac{3}{2}+\left(\dfrac{3}{2}\right)^2+...+\left(\dfrac{3}{2}\right)^{2023}\)
\(\dfrac{3}{2}C=\left(\dfrac{3}{2}\right)^2+\left(\dfrac{3}{2}\right)^3+...+\left(\dfrac{3}{2}\right)^{2024}\)
\(\dfrac{3}{2}C-C=\left(\dfrac{3}{2}\right)^2+\left(\dfrac{3}{2}\right)^3+...+\left(\dfrac{3}{2}\right)^{2024}-\dfrac{3}{2}-\left(\dfrac{3}{2}\right)^2-...-\left(\dfrac{3}{2}\right)^{2023}\)
\(\dfrac{1}{2}C=\left(\dfrac{3}{2}\right)^{2024}-\dfrac{3}{2}\)
\(C=2\left(\dfrac{3}{2}\right)^{2024}-3\)
\(\Rightarrow A=\dfrac{1}{2}+2\left(\dfrac{3}{2}\right)^{2024}-3\)
\(=2\left(\dfrac{3}{2}\right)^{2024}-\dfrac{5}{2}\)
\(\Rightarrow A-B=2\left(\dfrac{3}{2}\right)^{2024}-\dfrac{5}{2}-2\left(\dfrac{3}{2}\right)^{2024}=-\dfrac{5}{2}\)
H9
HT.Phong (9A5)
CTVHS
3 tháng 3
Xác suất thực nghiệm không phải mặt 4 chấm là:
\(\dfrac{40-13}{40}=\dfrac{27}{40}\)