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\(A=\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+100}\)
\(A=\frac{1}{3}+\frac{1}{6}+...+\frac{1}{5050}\)
\(A=2\left(\frac{1}{6}+\frac{1}{12}+...+\frac{1}{10100}\right)\)
\(A=2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{100.101}\right)=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{100}-\frac{1}{101}\right)\)
\(A=2.\left(\frac{1}{2}-\frac{1}{101}\right)\)
Tự tính
\(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{5050}\)
\(=2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{10100}\right)\)
\(=2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{100.101}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{100}-\frac{1}{101}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{101}\right)\)
\(=2.\frac{99}{202}\)
\(=\frac{99}{101}\)
Chào bạn, bạn hãy theo dõi câu trả lời của mình nhé!
Theo mình thì đề phải là \(A=3+\frac{3}{1+2}+\frac{3}{1+2+3}+\frac{3}{1+2+3+4}+...+\frac{3}{1+2+3+...+100}\).
Ta có :
\(A=3+\frac{3}{1+2}+\frac{3}{1+2+3}+\frac{3}{1+2+3+4}+...+\frac{3}{1+2+3+...+100}\)
\(=>A=3\left(1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+...+100}\right)\)
Đặt \(1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+...+100}\) là B. Ta có :
\(B=\)\(1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+...+100}\)
\(=>B=\frac{1}{1}+\frac{1}{\left(1+2\right)\cdot2:2}+\frac{1}{\left(1+3\right)\cdot3:2}+\frac{1}{\left(1+4\right)\cdot4:2}+...+\frac{1}{\left(1+100\right)\cdot100:2}\)
\(=>B=\frac{1}{1}+\frac{1}{3\cdot2:2}+\frac{1}{4\cdot3:2}+\frac{1}{5\cdot4:2}+...+\frac{1}{101\cdot100:2}\)
\(=>B=\frac{2}{1\cdot2}+\frac{2}{2\cdot3}+\frac{2}{3\cdot4}+\frac{2}{4\cdot5}+...+\frac{2}{100\cdot101}\)
\(=>B=2\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{100\cdot101}\right)\)
\(=>B=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{100}-\frac{1}{101}\right)\)
\(=>B=2\left(1-\frac{1}{101}\right)\)
\(=>B=2\cdot\frac{100}{101}=\frac{200}{101}\)
\(=>A=3B=3\cdot\frac{200}{101}=\frac{600}{101}\)
Chúc bạn học tốt!
\(A=1+\frac{1}{2}.\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+\frac{1}{4}\left(1+2+3+4\right)+...+\frac{1}{100}\left(1+2+3+...+100\right)\)
\(=1+\frac{1}{2}.\frac{2.3}{2}+\frac{1}{3}.\frac{3.4}{2}+\frac{1}{4}.\frac{4.5}{2}+...+\frac{1}{100}.\frac{100.101}{2}\)
\(=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+...+\frac{101}{2}=\frac{2+3+4+...+101}{2}=\frac{5151}{2}\)
\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2019^2}\)\(=\frac{1}{2.2}+\frac{1}{3.3}+...+\frac{1}{2019.2019}\)
\(=\frac{1}{2}-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}+...+\frac{1}{2019}-\frac{1}{2019}\)
\(=0\)
\(a)\) Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2010^2}\) ta có :
\(A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2009.2010}\)
\(A< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2009}-\frac{1}{2010}\)
\(A< 1-\frac{1}{2010}=\frac{2009}{2010}< 1\)
\(\Rightarrow\)\(A< 1\) ( đpcm )
Vậy \(A< 1\)
Chúc bạn học tốt ~
bài làm
C=1+3+32+.............+3100
C=3C−C2
3C=3+32+33+.............+399+3100+3101
C=1+3+32+..................+399+3100
3C-C=(3+32+33+.............+399+3100+3101)-(1+3+32+..................+399+3100)
Triệt tiêu các số hạng co giá trị tuyệt đối bằng nhau, ta được:
2C=-1+3100
⇒C=3100−12
D=2/D+D/3
2D=2101-2100+299-298+..............+23-22
D=2100-299+298-297+............+22-2
2D+D=2101-2100+299-298+..............+23-22+2100-299+298-297+............+22-2
Triệt tiêu các số hạng có giá trị tuyệt đối bằng nhau, ta được:
3D=2101-2
⇒D=2101−23
B=31×4 +54×9 +79×16 +.........+1981×100
Quan sát biểu thức, ta có nhận xét:
4-1=3;
9-4=5;
16-9=7;
.......;100-81=19
=> Hiệu hai số ở mẫu bằng giá trị ở tử
⇒B=1−14 +14 −19 +19 −116 +.......+181 −1100
⇒B=1−1/100
B=99/100 <100/100
Vậy B<1
\(S=\frac{1}{100}-\frac{2}{100}+\frac{3}{100}-...-\frac{98}{100}+\frac{99}{100}-\frac{100}{100}\)
\(=\frac{1-2+3-...-98+99-100}{100}\)
\(=\frac{\left[\left(1-2\right)+\left(3-4\right)+...+\left(97-98\right)+\left(99-100\right)\right]}{100}\)
\(=\frac{-1-1-1-...-1}{100}=\frac{-1.50}{100}=\frac{-50}{100}=\frac{-1}{2}\)
Vậy S=\(\frac{-1}{2}\)
\(S=\frac{1}{100}-\frac{2}{100}+\frac{3}{100}-\frac{4}{100}+\frac{5}{100}-...-\frac{98}{100}+\frac{99}{100}\)
\(S=\frac{\left(1-2\right)+\left(3-4\right)+\left(5-6\right)+....+\left(97-98\right)+\left(99-100\right)}{100}\)
\(S=\frac{-1+\left(-1\right)+\left(-1\right)+.....+\left(-1\right)+\left(-1\right)}{100}\)
Từ 1 đến 100 có 100 số số hạng => Có 50 cặp => có 50 số (-1)
=> \(S=\frac{50\cdot\left(-1\right)}{100}=\frac{-50}{100}=\frac{-1}{20}\)
Ta có : \(A=3+\frac{3}{1+2}+\frac{3}{1+2+3}+...+\frac{3}{1+2+...+100}\)
\(A=3\left(1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+...+100}\right)\)
Mà \(1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+...+100}=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{100.101}\)
\(=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{100}-\frac{1}{101}\right)=2\left(1-\frac{1}{101}\right)=\frac{200}{101}\)
\(\Rightarrow A=3.\frac{200}{101}=\frac{600}{101}\)