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\(A=\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+....+\frac{19}{9^2.10^2}\)
\(A=\frac{3}{1.4}+\frac{5}{4.9}+\frac{7}{9.16}+....+\frac{19}{81.100}\)
\(A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+....+\frac{1}{81}-\frac{1}{100}\)
\(A=1-\frac{1}{100}=\frac{99}{100}< 1\)
\(\Rightarrow A< 1\text{(đpcm) }\)
sửa đề câu 1 :
\(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{99}{100!}\)
\(=\frac{2-1}{2!}+\frac{3-1}{3!}+\frac{4-1}{4!}+...+\frac{100-1}{100!}\)
\(=\frac{1}{1!}-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{3!}-\frac{1}{4!}+...+\frac{1}{99!}-\frac{1}{100!}\)
\(=1-\frac{1}{100!}< 1\)
sửa đề câu 2
\(\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+\frac{3.4-1}{4!}+...+\frac{99.100-1}{100!}\)
\(=\frac{1.2}{2!}-\frac{1}{2!}+\frac{2.3}{3!}-\frac{1}{3!}+\frac{3.4}{4!}-\frac{1}{4!}+...+\frac{99.100}{100!}-\frac{1}{100!}\)
\(=\left(\frac{1.2}{2!}+\frac{2.3}{3!}+\frac{3.4}{4!}+...+\frac{99.100}{100!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{100!}\right)\)
\(=\left(1+1+\frac{1}{2!}+...+\frac{1}{98!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{100!}\right)\)
\(=2-\frac{1}{99!}-\frac{1}{100!}< 2\)
Phần C đề thiếu
\(D=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}\)
\(\Rightarrow3D=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}\)
\(\Rightarrow3D-D=(1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}})-\)\((\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}})\)
\(\Rightarrow2D=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
\(\Rightarrow6D=3+1+\frac{1}{3}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\)
\(\Rightarrow6D-2D=3-\frac{101}{3^{99}}+\frac{100}{3^{100}}\)
\(\Rightarrow4D=3-\frac{203}{3^{100}}\)
\(\Rightarrow D=\frac{3}{4}-\frac{\frac{203}{3^{100}}}{4}< \frac{3}{4}\left(đpcm\right)\)
#)Giải :
Bài 1 :
\(C=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\Leftrightarrow3C=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{100}}\)
\(\Leftrightarrow3C-C=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{100}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\right)\)
\(\Leftrightarrow2C=1-\frac{1}{3^{100}}\Leftrightarrow C=\frac{1-\frac{1}{3^{100}}}{2}< \frac{1}{2}\Rightarrow C< \frac{1}{2}\left(đpcm\right)\)
Bài 2 :
\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}=\frac{3}{1.4}+\frac{5}{4.9}+\frac{7}{9.16}+...+\frac{19}{81.100}\)
\(=\left(1-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{9}\right)+\left(\frac{1}{9}-\frac{1}{16}\right)+...+\left(\frac{1}{81}-\frac{1}{100}\right)=1-\frac{1}{100}=\frac{99}{100}< 1\)
\(\Rightarrow\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}< 1\left(đpcm\right)\)
Giúp mình nha. Bài cuối cùng của đề toán dài 36 bài của mình đó
\(A=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{100.100}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
Mà \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}< 1\)
Nên từ đây => \(A< 1\) (ĐPCM)
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)
Ta có : \(\frac{1}{2^2}=\frac{1}{2\cdot2}< \frac{1}{1\cdot2}\)
\(\frac{1}{3^2}=\frac{1}{3\cdot3}< \frac{1}{2\cdot3}\)
\(\frac{1}{4^2}=\frac{1}{4\cdot4}< \frac{1}{3\cdot4}\)
...
\(\frac{1}{100^2}=\frac{1}{100\cdot100}< \frac{1}{99\cdot100}\)
Cộng vế theo vế
=> \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)
=> \(A< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
=> \(A< \frac{1}{1}-\frac{1}{100}=\frac{99}{100}\)(1)
Lại có \(\frac{99}{100}< 1\)(2)
Từ (1) và (2) => \(A< \frac{99}{100}< 1\Rightarrow A< 1\left(đpcm\right)\)
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{100.100}\)
MÀ \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}>\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{100.100}\)
\(\Leftrightarrow\frac{1}{1}-\frac{1}{100}>\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{100.100}\)
\(\Leftrightarrow\frac{99}{100}>\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{100.100}\)
MÀ \(1>\frac{99}{100}\)
\(\Rightarrow1>\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{100.100}\)
\(\Rightarrow1>\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)
\(\Rightarrow A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< 1\left(đpcm\right)\)
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< 1\)
Ta có:\(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};...;\frac{1}{100^2}< \frac{1}{99.100}\)
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)(1)
Ta có:\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{1}-\frac{1}{100}=\frac{99}{100}< 1\)(2)
Từ (1) và (2)=>\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< 1\left(đpcm\right)\)
Thanh kiu mí bạn
Ta có : B = \(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{101}{3^{101}}\)(1)
=> 3B = \(1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{101}{3^{100}}\)(2)
Lấy (2) trừ (1) theo vế ta có :
3B - B = \(\left(1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{101}{3^{100}}\right)-\left(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}+\frac{101}{3^{101}}\right)\)
2B = \(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{100}}-\frac{101}{3^{101}}\)
Đặt C = \(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{100}}\)(3)
=> 3C = \(3+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\)(4)
Lấy (4) trừ (3) theo vế ta có
3C - C = \(\left(3+1+\frac{1}{3}+...+\frac{1}{3^{99}}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{100}}\right)\)
=> 2C = \(3-\frac{1}{3^{100}}\)
=> C = \(\frac{3}{2}-\frac{1}{3^{100}.2}\)
Khi đó 2B = \(\frac{3}{2}-\frac{1}{3^{100}.2}-\frac{101}{3^{101}}\)
=> B = \(\frac{3}{4}-\frac{1}{3^{100}.4}-\frac{101}{3^{101}.2}< \frac{3}{4}\left(\text{ĐPCM}\right)\)
Vậy B < 3/4
\(A=\frac{1}{2^2}+\frac{1}{3^3}+\frac{1}{4^2}+...+\frac{1}{100^2}< 1\)
Ta có : \(\frac{1}{2^2}< \frac{1}{1.2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}\)
\(\frac{1}{4^2}< \frac{1}{3.4}\)
....
\(\frac{1}{100^2}< \frac{1}{99.100}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}< 1\)
Nên \(A< 1\)