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=1(2+1)+2(3+1)+3(4+1)+...+100(101+1)
=1.2+1+2.3+2+3.4+3+...+100.101+100
=(1.2+2.3+3.4+..+100.101)+(1+2+3+...+100)
=333300+5000
=338300
Ta có :
\(A=\frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+\frac{1}{4.6}+...+\frac{1}{97.99}+\frac{1}{98.100}\)
\(A=\frac{1}{2}.\left(1-\frac{1}{3}\right)+\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}\right)+\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}\right)+\frac{1}{2}.\left(\frac{1}{4}-\frac{1}{6}\right)+...+\frac{1}{2}.\left(\frac{1}{97}-\frac{1}{99}\right)+\frac{1}{2}.\left(\frac{1}{98}-\frac{1}{100}\right)\)
\(A=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{97}-\frac{1}{99}+\frac{1}{98}-\frac{1}{100}\right)\)
\(A=\frac{1}{2}.\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{97}+\frac{1}{98}-\frac{1}{3}-\frac{1}{4}-\frac{1}{5}-\frac{1}{6}-...-\frac{1}{99}-\frac{1}{100}\right)\)
\(A=\frac{1}{2}.\left(1+\frac{1}{2}-\frac{1}{99}-\frac{1}{100}\right)< \frac{1}{2}.\left(1+\frac{1}{2}\right)=\frac{3}{4}\)
a) Số số hạng của dãy A là: (2020-5):2+1 = 404 (số)
Tổng A là: (2020+5)x404:2=409050
b) \(B=\frac{2}{1\times3}+\frac{2}{3\times5}+....+\frac{2}{99\times101}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{99}-\frac{1}{101}\)
\(=1-\frac{1}{101}=\frac{100}{101}\)
c) \(C=\frac{1}{2\times4}+\frac{1}{4\times6}+\frac{1}{6\times8}+...+\frac{1}{98\times100}\)
\(=\frac{1}{2}\times\left(\frac{2}{2\times4}+\frac{2}{4\times6}+\frac{2}{6\times8}+....+\frac{2}{98\times100}\right)\)
\(=\frac{1}{2}\times\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+....+\frac{1}{98}-\frac{1}{100}\right)\)
\(=\frac{1}{2}\times\left(1-\frac{1}{100}\right)=\frac{1}{2}\times\frac{99}{100}=\frac{99}{200}\)
Vậy .....
A = 5 + 10 + 15 + ... + 2015 + 2020
Số số hạng là : 404
A = 409050
\(B=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{99\cdot101}\)
\(B=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(B=1-\frac{1}{101}=\frac{101-1}{101}=\frac{100}{101}\)
\(C=\frac{1}{2\cdot4}+\frac{1}{4\cdot6}+\frac{1}{6\cdot8}+...+\frac{1}{98\cdot100}\)
\(C=\frac{1}{2}\cdot\left(\frac{1}{2}-\frac{1}{4}\right)+\frac{1}{2}\cdot\left(\frac{1}{4}-\frac{1}{6}\right)+\frac{1}{2}\cdot\left(\frac{1}{6}-\frac{1}{8}\right)+...+\frac{1}{2}\cdot\left(\frac{1}{98}-\frac{1}{100}\right)\)
\(C=\frac{1}{2}\cdot\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{98}-\frac{1}{100}\right)\)
\(C=\frac{1}{2}\cdot\left(\frac{1}{2}-\frac{1}{100}\right)=\frac{1}{2}\cdot\frac{49}{100}=\frac{49}{200}\)
a) \(\frac{1}{1x3}+\frac{1}{3x5}+\frac{1}{5x7}+...+\frac{1}{2007x2009}\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2007}-\frac{1}{2009}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{2009}\right)=\frac{1}{2}\cdot\frac{2008}{2009}=\frac{1004}{2009}\)
....
các bài cn lại bn lm tương tự nha
B=\(\frac{4}{1\cdot3}\cdot\frac{9}{2\cdot4}\cdot\cdot\cdot\cdot\cdot\frac{100}{9\cdot11}\)
=\(\frac{2\cdot2}{1\cdot3}\cdot\frac{3\cdot3}{2\cdot4}\cdot\cdot\cdot\cdot\cdot\frac{10\cdot10}{9\cdot11}\)
Sau khi rút gọn còn:
\(\frac{2}{1}\cdot\frac{10}{11}=\frac{20}{11}\)
\(B=\frac{4}{1\cdot3}\cdot\frac{9}{2\cdot4}\cdot\frac{16}{3\cdot5}\cdot...\cdot\frac{100}{9\cdot10}\)
\(B=\frac{2\cdot2}{1.3}\cdot\frac{3\cdot3}{2\cdot4}\cdot\frac{4\cdot4}{3\cdot5}\cdot...\cdot\frac{10\cdot10}{9\cdot10}\)
\(B=\frac{\left(2\cdot3\cdot4\cdot...\cdot10\right)\cdot\left(2\cdot3\cdot4\cdot...\cdot10\right)}{\left(1\cdot2\cdot3\cdot...\cdot9\right)\cdot\left(3\cdot4\cdot5\cdot...\cdot10\right)}\)
\(B=10\cdot2\)
\(B=20\)
S = 2*4+4*6+6*8+...+46*48+48*50
S6 = 2*4*6+4*6*6+6*8*6+........................+46*48*6+48*50*6
S6=2*4*(6-0)+4*6*(8-2)+6*8*(10-4)+.................................+46*48*(50-44)+48*50*(52-46)
S6 = 2*4*6+4*6*8-2*4*6+6*8*10-4*6*8+..........................................+46*48*50-44*46*48+48*50*52-46*48*50
S6 = 48*50*52=124800
S=124800/6=20800
\(S=2\cdot4+4\cdot6+...+48\cdot50\)
\(S=2\left(1\cdot2+2\cdot3+...+24\cdot25\right)\)
\(\Rightarrow3S=2\left(1\cdot2\left(3-0\right)+2\cdot3\left(4-1\right)+...+24\cdot25\left(26-23\right)\right)\)
\(\Rightarrow3S=2\left(1\cdot2\cdot3-0\cdot1\cdot2+2\cdot3\cdot4-1\cdot2\cdot3+...+24\cdot25\cdot26-23\cdot24\cdot25\right)\)
\(\Rightarrow3S=2\cdot24\cdot25\cdot26\)
\(\Rightarrow S=2\cdot8\cdot25\cdot26=10400\)
\(\Rightarrow6S=10400\cdot6=62400\)
Đáp án =2525 vì câu của cậu có người hỏi rồi