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a)
\(A=\dfrac{3}{4}.\dfrac{8}{9}...\dfrac{9999}{10000}\)
\(=\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}...\dfrac{99.101}{100.100}\)
\(=\dfrac{1.2...99}{2.3...100}.\dfrac{3.4...101}{2.3...100}\)
\(=\dfrac{1}{100}.\dfrac{101}{2}\)
\(=\dfrac{101}{200}\)
A=1.2.3+2.3.4+....+99.100.101
4A=1.2.3.4+2.3.4.(5-1)+3.4.5.(6-2)+....+98.99.100.(101-97)
4A=1.2.3.4+2.3.4.5-1.2.3.4+3.4.5.6-3.4.5.2+....+98.99.100.101-98.99.100.97
4A=98.99.100.101
4A=97990200
A=97990200/4
A=24497550
B=1.2+3.4+5.6+7.8+8.9+...+999.1000
3B=1.2.3+2.3.(4-1)+3.4(5-2)+....+998.999(1001-998)
3B=1.2.3+2.3.4-2.3.1+3.4.5-3.4.2+....+998.999.1001-998.999.998
3B=999.1000.1001
3B=999999000
B=999999000/3
B=333333000
C=1+4+9+16+25+36+.....+10000
C=1^2+2^2+3^2+4^2+5^2+6^2+....+100^2
C=(1^2+3^2+5^2+.....+99^2)+(2^2+4^2+6^2+....+100^2)
C=99.100.101/6 + 100.101.102/6
C=166650 +171700
C=338350
Còn câu d bạn dựa vào câu c là làm được ngay bây h mk mỏi tay rùi ko muốn đánh nữa khi nào rảnh mk gửi công thức cho nha bây h mk bận rùi.
chúc bn học tốt
A=1.2.3+2.3.4+....+99.100.101
4.A=1.2.3.(4-0)+2.3.4.(5-1)+...+99.100.101.(102-98)
4.A=1.2.3.1-0.1.2.3+2.3.4.5-1.2.3.4+....+99.100.101.102-98.99.100.101
4.A=99.100.101.102
A=\(\frac{99.100.101.102}{4}\)
B=1.2+2.3+3.4+...+999.1000
3.B=1.2.(3-0)+2.3.(4-1)+3.4.(5-2)+.....+999.1000.(1001-998)
3.B=1.2.3-0.1.2+2.3.4-1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+......+999.1000.1001-998.999.1000
3.B=999.1000.1001
=>B=\(\frac{999.1000.1001}{3}\)
C và D dễ lắm bạn tự làm nhé
\(A=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.....\frac{9999}{10000}\)
\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}......\frac{99.101}{100.100}\)
\(=\frac{1.2.3...99}{2.3.4....100}.\frac{3.4.5....101}{2.3.4....100}\)
\(=\frac{1}{100}.\frac{101}{2}\)
\(=\frac{101}{200}\)
A=\(\dfrac{3}{4}.\dfrac{8}{9}.....\dfrac{9999}{10000}\)
A=\(\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}.....\dfrac{99.101}{100.100}\)
A=\(\dfrac{1.2.3.....99}{2.3.4.....100}.\dfrac{3.4.....101}{2.3.4.....100}\)
A=\(\dfrac{1}{100}.\dfrac{101}{2}\)
A=\(\dfrac{101}{200}\)
\(A=\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}.\dfrac{3.5}{4.4}.....\dfrac{99.101}{100.100}\\ =\dfrac{1}{2}.\dfrac{101}{100}=\dfrac{101}{200}\)
\(B=\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{9}\right)...\left(1-\dfrac{1}{10000}\right)\\ =\dfrac{3}{4}.\dfrac{8}{9}...\dfrac{9999}{10000}\)
(làm như câu a)
\(A=\dfrac{3}{4}\cdot\dfrac{8}{9}\cdot\dfrac{15}{16}\cdot...\cdot\dfrac{9999}{10000}\\ =\dfrac{1\cdot3}{2\cdot2}\cdot\dfrac{2\cdot4}{3\cdot3}\cdot\dfrac{3\cdot5}{4\cdot4}\cdot...\cdot\dfrac{99\cdot101}{100\cdot100}\\ =\dfrac{1\cdot3\cdot2\cdot4\cdot3\cdot5\cdot...\cdot99\cdot101}{2\cdot2\cdot3\cdot3\cdot4\cdot4\cdot...\cdot100\cdot100}\\ =\dfrac{\left(1\cdot2\cdot3\cdot...\cdot99\right)\cdot\left(3\cdot4\cdot5\cdot...\cdot101\right)}{\left(2\cdot3\cdot4\cdot...\cdot100\right)\cdot\left(2\cdot3\cdot4\cdot...\cdot100\right)}\\ =\dfrac{1\cdot101}{100\cdot2}\\ =\dfrac{101}{200}\)
\(C=\left(1+\dfrac{1}{1\cdot3}\right)\cdot\left(1+\dfrac{1}{2\cdot4}\right)\cdot\left(1+\dfrac{1}{3\cdot5}\right)\cdot...\left(1+\dfrac{1}{99\cdot101}\right)\\ =\left(\dfrac{1\cdot3}{1\cdot3}+\dfrac{1}{1\cdot3}\right)\cdot\left(\dfrac{2\cdot4}{2\cdot4}+\dfrac{1}{2\cdot4}\right)\cdot\left(\dfrac{3\cdot5}{3\cdot5}+\dfrac{1}{3\cdot5}\right)\cdot...\cdot\left(\dfrac{99\cdot101}{99\cdot101}+\dfrac{1}{99\cdot101}\right)\\ =\left(\dfrac{2^2-1}{1\cdot3}+\dfrac{1}{1\cdot3}\right)\cdot\left(\dfrac{3^2-1}{2\cdot4}+\dfrac{1}{2\cdot4}\right)\cdot\left(\dfrac{4^2-1}{3\cdot5}+\dfrac{1}{3\cdot5}\right)\cdot...\cdot\left(\dfrac{100^2-1}{99\cdot101}+\dfrac{1}{99\cdot101}\right)\\ =\dfrac{2^2}{1\cdot3}\cdot\dfrac{3^2}{2\cdot4}\cdot\dfrac{4^2}{3\cdot5}\cdot...\cdot\dfrac{100^2}{99\cdot101}\\ =\dfrac{2^2\cdot3^2\cdot4^2\cdot...\cdot100^2}{1\cdot3\cdot2\cdot4\cdot3\cdot5\cdot...\cdot99\cdot101}\\ =\dfrac{\left(2\cdot3\cdot4\cdot...\cdot100\right)\cdot\left(2\cdot3\cdot4\cdot...\cdot100\right)}{\left(1\cdot2\cdot3\cdot...\cdot99\right)\cdot\left(3\cdot4\cdot5\cdot...\cdot101\right)}\\ =\dfrac{100\cdot2}{1\cdot101}=\dfrac{200}{101}\)