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Cho A=1/1.2 + 1/2.3 + + 1/ 3.4+...+1/49.50 ; B = 1.2+2.3+3.4+4.5+5.6+...+49.50
Tính 50 mủ 2 A – B/17
\(\frac{2}{1\cdot2}+\frac{2}{2\cdot3}+\frac{2}{3\cdot4}+........+\frac{2}{8\cdot9}+\frac{2}{9\cdot10}\)
\(=2\cdot\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+......+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}\right)\)
\(=2\cdot\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+........+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)
\(=2\cdot\left(\frac{1}{1}-\frac{1}{10}\right)\)
\(=2\cdot\frac{9}{10}=\frac{9}{5}\)
đúng nha !
Ta có:
2/1.2+2/2.3+2/3.4+2/4.5+2/5.6+2/6.7+2/7.8+2/8.9+2/9.10
=2.(1/1.2+1/2.3+1/3.4+1/4.5+1/5.6+1/6.7+1/7.8+1/8.9+1/9.10)
=2.(1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9+1/9-1/10)
=2.(1-1/10)
=2.9/10
=9/5
câu a) (a^2+2a+a+2)(a+3)-(a^2+a)(a+2)= (3a+3)(a+2)
suy ra: a^3+3x^2+2a^2+6a+a^2+3a+2a+6-a^3-2x^2-a^2-2a= 3a^2+6a+3a+6
3a^2+9a+6=3a^2+9a+6
câu b)
c) Đặt \(A=1\cdot2+2\cdot3+3\cdot4+...+99\cdot100\)
Ta có: \(A=1\cdot2+2\cdot3+3\cdot4+...+99\cdot100\)
\(\Leftrightarrow3A=3\cdot\left(1\cdot2+2\cdot3+3\cdot4+...+99\cdot100\right)\)
\(\Leftrightarrow3A=1\cdot2\cdot3+2\cdot3\cdot\left(4-1\right)+3\cdot4\cdot\left(5-2\right)+...+99\cdot100\cdot\left(101-98\right)\)
\(\Leftrightarrow3\cdot A=1\cdot2\cdot3-1\cdot2\cdot3+2\cdot3\cdot4-2\cdot3\cdot4+...+98\cdot99\cdot100-98\cdot99\cdot100+99\cdot100\cdot101\)
\(\Leftrightarrow3\cdot A=99\cdot100\cdot101\)
\(\Leftrightarrow A=33\cdot100\cdot101=333300\)
b) Ta có: \(1+2-3-4+...+97+98-99-100\)
\(=\left(1+2-3-4\right)+\left(5+6-7-8\right)+...+\left(97+98-99-100\right)\)
\(=\left(-4\right)+\left(-4\right)+...+\left(-4\right)\)
\(=-4\cdot25=-100\)
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
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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
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Q = 1² + 2² + 3² + ... + 100² + 101²
= 101.102.(2.101 + 1) : 6
= 348551
Vì
\(\frac{1}{2}< \frac{2}{3};\frac{3}{4}< \frac{4}{5};\frac{5}{6}< \frac{6}{7};...;\frac{99}{100}< \frac{100}{101}\)
\(\Rightarrow\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}< \frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{100}{101}\)
\(\Rightarrow a< b\)