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Bài 5:
a) Ta có: \(A=1\cdot2+2\cdot3+3\cdot4+...+9\cdot10\)
\(\Leftrightarrow3\cdot A=3\cdot\left(1\cdot2+2\cdot3+3\cdot4+...+9\cdot10\right)\)
\(\Leftrightarrow3A=1\cdot2\cdot\left(3-0\right)+2\cdot3\cdot\left(4-1\right)+3\cdot4\cdot\left(5-2\right)+...+9\cdot10\cdot\left(11-8\right)\)
\(\Leftrightarrow3A=1\cdot2\cdot3+2\cdot3\cdot4-1\cdot2\cdot3+3\cdot4\cdot5-2\cdot3\cdot4+...+8\cdot9\cdot10-8\cdot9\cdot10+9\cdot10\cdot11\)
\(\Leftrightarrow3\cdot A=9\cdot10\cdot11=90\cdot11=990\)
hay A=330
Vậy: A=330
Bài 1 Số số hạng của dãy là : (50-1):1+1=50(số hạng )
S = (50+1) x 50 : 2 = 1275
A = 1.2 + 2.3 + ... + 99.100
3A = 1.2.3 + 2.3.(4-1) + ... + 99.100.(101-98)
3A = 1.2.3 + 2.3.4 - 1.2.3 + ... + 99.100.101 - 98.99.100
3A = 99.100.101
3A = 999900
A = 333300
C = 1.2.3 + 2.3.4 + ... + 49.50.51
4C = 1.2.3.4 + 2.3.4.(4-1) + ... + 49.50.51.(52-48)
4c = 1.2.3.4 + 2.3.4.5 - 1.2.3.4 + ... + 49.50.51.52 - 48.49.50.51
4C = 49.50.51.52
4C = 6497400
C = 1624350
A=1.2+2.3+3.4+.............+2019.2020
3A=1.2.3+2.3.3+3.4.3+........................+2019.2020.3
3A=1.2.3+2.3.(4-1)+3.4.(5-2)+..............+2019.2020.(2021-2018)
3A=1.2.3-1.2.3+2.3.4-2.3.4+3.4.5-3.4.5+.............-2018.2019.2020+2019.2020.2021
3A=2019.2020.2021
A=2019.2020.2021 / 3
A=2747468660
Vậy A=2747468660 .
🎀
\(A = 1.2+2.3+3.4+4.5+...+99.100\)
\(3A= 1.2.3+2.3.3+3.4.3+4.5.3+\)\(...+\)
\(99.100.3\)
\(3A = 1.2.3+2.3.(4-1)+3.4. (5-2)+\)
\(4.5. (6-3)+...+99.100. (101-98)\)
\(3A = 1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+\)
\(4.5.6-3.4.5+...+99.100.101-98.99.100\)
\(3A = 99 .100 .101\)
\(A = 99 .100 . 101 ÷ 3 \)
\(A = 333300\)
A = 1.2 + 2.3 + 3.4 + ....... + 99.100
3A = 1.2.3 + 2.3.3 + 3.4.3 + ....... + 99 . 100 . 3
3A = 1.2.3 + 2.3.(4-1) + 3.4.(5-2) +.... + 99.100.(101-98)
3A = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ..... + 99 . 100 . 101 - 98 . 99 . 100
3A = (1.2.3 - 1.2.3) + (2.3.4-2.3.4) + ... + (98.99.100 - 98.99.100) + 99 . 100 . 101
3A = 99 . 100 . 101 = 999900
A = 999900 : 3 = 343400
# Học tốt☘️#
\(E=\frac{1}{1.2}-\frac{1}{1.2.3}+\frac{1}{2.3}-\frac{1}{2.3.4}+....+\frac{1}{99.100}-\frac{1}{99.100.101}\)
\(=\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)-\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{99.100.101}\right)\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
\(B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{99.100.101}\)
\(=\frac{1}{2}\left(\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+...+\frac{101-99}{99.100.101}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{99.100}-\frac{1}{100.101}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{100.101}\right)=\frac{5049}{20200}\)
Suy ra \(E=A-B=\frac{99}{100}-\frac{5049}{20200}=\frac{14949}{20200}\)
Câu 1:
Đặt S = 1.2+2.3+3.4+...+30.31
3 S = 1.2.3+2.3.3+3.4.3+...+30.31.3
3 S = 1.2.(3-0) + 2.3.(4-1) + 3.4.(5-2) + ...+ 30.31.(32-29)
3S = 1.2.3 + 2.3.4-2.3 + 3.4.5-2.3.4 + ...+ 30.31.32-29.30.31
3S= 30.31.32
S= 30.31.32/3
\(A=\frac{4}{1.2}+\frac{4}{2.3}+\frac{4}{3.4}+...+\frac{4}{2019.2020}\)
\(\frac{1}{4}A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\)
\(\frac{1}{4}A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\)
\(\frac{1}{4}A=1-\frac{1}{2020}=\frac{2019}{2020}\)
\(\Rightarrow A=\frac{2019}{2020}:\frac{1}{4}=\frac{2019}{505}\)
Vậy \(A=\frac{2019}{505}.\)
\(B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}\)
\(\Rightarrow2B=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{98.99.100}\)
\(2B=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{98.99}-\frac{1}{99.100}\)
\(2B=\frac{1}{1.2}-\frac{1}{99.100}=\frac{4949}{9900}\)
\(\Rightarrow B=\frac{4949}{9900}:2=\frac{4949}{19800}\)
Vậy \(B=\frac{4949}{19800}.\)
\(A=\frac{4}{1\cdot2}+\frac{4}{2\cdot3}+\frac{4}{3\cdot4}+...+\frac{4}{2019\cdot2020}\)
\(A=4\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2018\cdot2019}\right)\)
\(A=4\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2018}-\frac{1}{2019}\right)\)
\(A=4\left(1-\frac{1}{2019}\right)=4\cdot\frac{2018}{2019}\)
Đến đây tự tính
\(B=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{98\cdot99\cdot100}\)
\(B=\frac{1}{2}\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+...+\frac{2}{98\cdot99\cdot100}\right)\)
\(B=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{98\cdot99}-\frac{1}{99\cdot100}\right)\)
\(B=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{99\cdot100}\right)=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{9900}\right)\)
Số hơi bị dữ nên tính nốt nhé