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Cau a) 1/1.2 +1/2.3 +1/3.4+...+1/99.100= 1/1-1/2+1/2-1/3+...+1/99-1/100
=1/1-1/100=99/100
99/100<1 thì 1/1.2 +1/2.3+1/3.4+...+1/99.100<1
Câu b): Ta có: 1/2^2<1/1.2
1/3^2<1/2.3
...............(so sánh như vậy với các số khác)
1/2016^2<1/2015.2016
Áp dụng của câu a ta thêm vào sau về thành: 1/1.2+1/2.3+1/3.4+...+1/2015.2016
=1/1-1/2+1/2-1/3+1/3-1/4+...+1/2015-1/2016
=1/1-1/2016
=2015/2016<1
Ma :1/2^2+1/3^2+1/4^2+...+1/2016^2<1/1.1+1/2.3+1/3.4+...+1/2015.2016
Nen:1/1^2+1/3^2+1/4^2+...+1/2016^2<1
\(\dfrac{4}{3.7}+\dfrac{4}{7.11}+...+\dfrac{4}{23.27}\)
= \(4.\left(\text{}\text{}\text{}\text{}\text{}\text{}\dfrac{4}{3.7}+\dfrac{4}{7.11}+...+\dfrac{4}{23.27}\right)\)
=\(1.\left(\dfrac{1}{3.7}+\dfrac{1}{7.11}+...+\dfrac{1}{23.27}\right)\)
= \(1.\left(\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+...+\dfrac{1}{23}-\dfrac{1}{27}\right)\)
=\(1.\left(\dfrac{1}{3}-\dfrac{1}{27}\right)\)
=\(1.\left(\dfrac{9}{27}-\dfrac{1}{27}\right)\)
= \(1.\dfrac{8}{27}\)
= \(\dfrac{8}{27}\)
Ta có ; K = \(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.....+\frac{1}{45}\)
\(=1+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{90}\)
\(=1+\left(\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+.....+\frac{2}{9.10}\right)\)
\(=1+2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.....+\frac{1}{9.10}\right)\)
\(=1+2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{9}-\frac{1}{10}\right)\)
\(=1+2\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(=1+1-\frac{1}{5}\)(nhân phá ngoặc)
\(=2-\frac{1}{5}\)< 2
Vậy K = \(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.....+\frac{1}{45}\)< 2
Tính tổng:a)3+3/5+3/25+3/125+3/625
b)M=4/3.7+4/7.11+4/11.15+...+8/95.99
c)N=1/2+1/6+1/12+1/20+...+1/90
Ta có : \(M=\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+.....+\frac{4}{95.99}\)
\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+......+\frac{1}{95}-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}\)
\(=\frac{32}{99}\)
\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2021.2022}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2021}-\dfrac{1}{2022}\)
\(=1-\dfrac{1}{2022}=\dfrac{2021}{2022}\)
\(B=\dfrac{4}{3.7}+\dfrac{4}{7.11}+\dfrac{4}{11.15}+...+\dfrac{4}{107.111}\)
\(=\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{15}+...+\dfrac{1}{107}-\dfrac{1}{111}\)
\(=\dfrac{1}{3}-\dfrac{1}{111}=\dfrac{12}{37}\)
A = SCSH: ( 102 - 1 ) : 1 + 1 = 102
A = Tổng: ( 102 + 1 ) . 102 : 2 = 5253
Vậy KQ là: 5253
B = SCSH: ( 2998 - 1 ) : 3 + 1 = 1000
B = Tổng: ( 2998 + 1 ) . 1000 : 2 = 1499500
Vậy KQ là 1499500
mk làm phần a thui nhé
a. A = 1/2.3 + 1/3.4 + 1/4.5 + 1/5.6
A = 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6
A = 1/2 - 1/6
A= 3/6 - 1/6
A = 1/3
\(B=\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot14}\)
\(b=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}\)
\(b=\frac{1}{2}-\frac{1}{14}\)
\(b=\frac{3}{7}\)
\(d=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{110}\)
\(d=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{10\cdot11}\)
\(d=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\)
\(d=1-\frac{1}{11}\)
\(d=\frac{10}{11}\)
\(e=\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+\frac{1}{154}+\frac{1}{238}+\frac{1}{340}\)
\(e=\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+\frac{1}{11\cdot14}+\frac{1}{14\cdot17}+\frac{1}{17\cdot20}\)
\(e=\frac{1}{3}\left(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{17\cdot20}\right)\)
\(e=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{17}-\frac{1}{20}\right)\)
\(e=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{20}\right)\)
\(e=\frac{1}{3}\cdot\frac{9}{20}=\frac{3}{20}\)