(1-2/2.3)(1-2/3.4)(1-2/4.5)...(1-2/99.100)
Cho A= 1/2.3+1/3.4+1/4.5+...+1/99.100. So sánh A với 1/2
\(A=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{99\cdot100}\)
\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-...+\frac{1}{99}-\frac{1}{100}\)
\(A=\frac{1}{2}-\frac{1}{100}\)
\(\frac{1}{2}-\frac{1}{100}< \frac{1}{2}\)
\(\Rightarrow A< \frac{1}{2}\)
\(A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=\frac{1}{2}-\frac{1}{100}\)
\(\Leftrightarrow A< \frac{1}{2}\)
\(A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=\frac{1}{2}-\frac{1}{100}\)
\(A=\frac{49}{100}\)
Vậy A < 1/2.Vì 1/2 = 50/100 nên 49/100 > 50/100 nên A > 1/2
Tính nhanh :
A=1/30 + 1/42 + 1/56 + 1/72 + ... + 1/210
B=2/2.3 + 2/3.4 + 2/4.5 + ... + 2/99.100
\(A=\frac{1}{30}+\frac{1}{42}+...+\frac{1}{210}\)
\(A=\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{14.15}\)
\(A=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{14}-\frac{1}{15}\)
\(A=\frac{1}{5}-\frac{1}{15}\)
Tự tính nha :)
\(B=\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{99.100}\)
\(B=2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)
\(B=2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)
\(B=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(B=2\left(\frac{1}{2}-\frac{1}{100}\right)\)
Tự làm
\(A=\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+....+\frac{1}{210}\)
\(\Rightarrow A=\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+....+\frac{1}{14\cdot15}\)
\(\Rightarrow A=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+......+\frac{1}{14}-\frac{1}{15}\)
\(\Rightarrow A=\frac{1}{5}-\frac{1}{15}\)
\(\Rightarrow A=\frac{2}{15}\)
S=1+2+2^2+2^3+2^4+...+2^100
S=1.2+2.3+3.4+4.5+...+99.100+100.101
Q=1^2+2^2+3^2+...+100^2+101^2
S = 1 + 2 + 2² + 2³ + 2⁴ + ... + 2¹⁰⁰
2S = 2 + 2² + 2³ + 2⁴ + ... + 2¹⁰¹
S = 2S - S
= (2 + 2² + 2³ + ... + 2¹⁰¹) - (1 + 2 + 2² + ... + 2¹⁰⁰)
= 2¹⁰¹ - 1
------------
S = 1.2 + 2.3 + 3.4 + ... + 99.100 + 100.101
3S = 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 99.100.(101 - 98) + 100.101.(102 - 99)
= 1.2.3 - 1.2.3 + 2
3.4 - 2.3.4 + 3.4.5 - ... - 98.99.100 + 99.100.101 - 99.100.101 + 100.101.102
= 100.101.102
S = 100 . 101 . 102 : 3
= 343400
------------
Q = 1² + 2² + 3² + ... + 100² + 101²
= 101.102.(2.101 + 1) : 6
= 348551
Câu 1: 1/2.3+1/3.4+1/4.5+.....+1/99.100
Câu 2:: tính tổng
A=2 mũ 0+2 mũ 1+2 mũ 2+.......+ 2 mũ 2010
Câu 1
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=\frac{1}{2}-\frac{1}{99}=\frac{49}{100}\)
cho mình nha bạn
Thieu Gia Ho Hoang k tl thi dung kiem l...i....k...e
\(\left(1-\frac{2}{2.3}\right)\left(1-\frac{2}{3.4}\right)\left(1-\frac{2}{4.5}\right)......\left(1-\frac{2}{99.100}\right)\)
Tính nhanh:
1)1.2+2.3+3.4+4.5+...+99.100
2) 1:20+1:44+1:77+1:119+1:170
Bài 1 :
Đặt A=1.2+2.3+3.4+4.5+.........+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
A=33.100.101
A=333300
Bài 2 :
1:20 + 1:44 + 1:77 + 1:119 + 1:170 = \(\frac{1}{20}+\frac{1}{44}+\frac{1}{77}+\frac{1}{119}+\frac{1}{170}=\frac{1}{10}=0,1\)
1)1.2+2.3+3.4+4.5+...+99.100
đặt 3D=1.2+2.3+3.4+...+99.100
=1.2.3+2.8.3+...+3.4.3+4.5.3+...+99.100.3
=1.2.3+2.3.(4-1)+3.4.(5-2)+4.5.(6-3)+...+99.100.(101-98)
=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+4.5.6-3.4.5
=99.100.101
=999900
D=999900:3=333300
nếu đúng nhớ cảm ơn nhak. mình ko bít làm bài 2
\(\frac{1}{20}+\left(\frac{1}{44}+\frac{1}{77}\right)+\left(\frac{1}{119}+\frac{1}{170}\right)=\frac{1}{20}+\left(\frac{1}{11}.\frac{1}{4}+\frac{1}{11}.\frac{1}{7}\right)+\left(\frac{1}{17}.\frac{1}{7}+\frac{1}{17}.\frac{1}{10}\right)\)
= \(\frac{1}{20}+\frac{1}{11}.\left(\frac{1}{4}+\frac{1}{7}\right)+\frac{1}{17}.\left(\frac{1}{7}+\frac{1}{10}\right)=\frac{1}{20}+\frac{1}{11}.\frac{11}{28}+\frac{1}{17}.\frac{17}{70}=\frac{1}{20}+\frac{1}{28}+\frac{1}{70}\)
= \(\frac{1}{20}+\frac{1}{14}.\left(\frac{1}{2}+\frac{1}{5}\right)=\frac{1}{20}+\frac{1}{14}.\frac{7}{10}=\frac{1}{20}+\frac{1}{20}=\frac{2}{20}=0,1\)
\(B=\left(1-\frac{2}{2.3}\right).\left(1-\frac{2}{3.4}\right).\left(1-\frac{2}{4.5}\right)....\left(1-\frac{2}{99.100}\right)\)
Chứng tỏ rằng: \(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{99.100}< \dfrac{1}{2}\)
Lời giải:
Ta có:
\(A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(A=\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{100-99}{99.100}\)
\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{99}-\frac{1}{100}\)
\(A=\frac{1}{2}-\frac{1}{100}< \frac{1}{2}\)
Vậy ta có đpcm.
tính
\(\left(1-\frac{2}{2.3}\right).\left(1-\frac{2}{3.4}\right).\left(1-\frac{2}{4.5}\right)...\left(1-\frac{2}{99.100}\right)\)
\(A=\left(1-\frac{2}{2\cdot3}\right)\cdot\left(1-\frac{2}{3\cdot4}\right)\cdot\left(1-\frac{2}{4\cdot5}\right)\cdot...\cdot1-\frac{2}{99\cdot100}\)
\(2A=1-\left(\frac{1}{2\cdot3}\cdot\frac{1}{3\cdot4}\cdot\frac{1}{4\cdot5}\cdot...\cdot\frac{1}{99\cdot100}\right)\)
\(2A=1-\left(\frac{1}{2}-\frac{1}{3}\cdot\frac{1}{3}-\frac{1}{4}\cdot\frac{1}{4}-\frac{1}{5}\cdot...\cdot\frac{1}{99}\cdot\frac{1}{100}\right)\)
\(2A=1-\left(\frac{1}{2}-\frac{1}{100}\right)\)
\(2A=1-\frac{49}{100}\)
\(2A=\frac{51}{100}\)
\(A=\frac{51}{100}:2\)
\(A=\frac{51}{200}\)