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a, Ta có : \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{199.200}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{199}-\frac{1}{200}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{199}+\frac{1}{200}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{199}+\frac{1}{200}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)
=> \(\frac{\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{199.200}}{\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}}=1\)
=> đpcm
Study well ! >_<
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
\(M=\frac{1}{2}.\frac{3}{4}.\frac{4}{5}...\frac{99}{100}\)
\(\Leftrightarrow M=\frac{1}{2}.\frac{3.4...99}{4.5...100}\)
\(\Leftrightarrow M=\frac{1}{2}.\frac{3}{100}\)
\(\Leftrightarrow M=\frac{3}{200}\)
\(N=\frac{2}{3}.\frac{4}{5}.\frac{5}{6}...\frac{100}{101}\)
\(\Leftrightarrow N=\frac{2}{3}.\frac{4.5...100}{5.6...101}\)
\(\Leftrightarrow N=\frac{2}{3}.\frac{4}{101}\)
\(\Leftrightarrow N=\frac{8}{303}\)
101 + 100 + ... + 2 + 1 = 101x102/2 = 101x51 = 5151
101 - 100 + 99 - .. + 1 = ( 101 -100 ) + ( 99 - 98 ) + ... + ( 3 - 2 ) + 1 = 1 + 1 + 1 + ... + 1 ( 51 số ) = 51
suy ra C = 5151/51 = 101
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3737x43 - 4343x36 = 37x101x43 - 43x101x36 = 43x101 = 4343
2 + 4 + 6 +... + 100 = 2x( 1 + 2 + ... + 50 ) = 2x50x51/2 = 50x51 = 2550
vậy D = 4343/2550
Vì
\(\frac{1}{2}< \frac{2}{3};\frac{3}{4}< \frac{4}{5};\frac{5}{6}< \frac{6}{7};...;\frac{99}{100}< \frac{100}{101}\)
\(\Rightarrow\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}< \frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{100}{101}\)
\(\Rightarrow a< b\)
*) A=1+6+11+16+21+....+101
Dãy trên có: \(\left(101-1\right):5+1=21\)(số số hạng)
\(\Rightarrow A=\frac{\left(101+1\right)\cdot21}{2}=1071\)
*) Đặt C=\(1^2+2^2+3^2+....+98^2=1\cdot1+2\cdot2+3\cdot3+....+98\cdot98\)
\(\Rightarrow B-C=\left(1\cdot2+2\cdot3+3\cdot4+....+98\cdot99\right)-\left(1\cdot1+2\cdot2+3\cdot3+....+98\cdot98\right)\)
\(=\left(1\cdot2-1\cdot1\right)+\left(2\cdot3-2\cdot2\right)+\left(3\cdot4-3\cdot3\right)+.....+\left(98\cdot99-98\cdot98\right)\)
\(=1\left(2-1\right)+2\left(3-2\right)+3\left(4-3\right)+....+98\left(99-1\right)\)
\(=1\cdot1+2\cdot1+3\cdot1+....+98\cdot1\)
\(=1+2+3+....+98\)
\(=\frac{\left(98+1\right)\cdot98}{2}=4851\)
A = 1 + 6 + 11 + 16 +21 +... + 101
Số chữ số của tổng A là :
( 101 - 1 ) : 5 + 1 = 21 (số)
Tổng A = 1 + 6 + ... + 101 = (101 + 1) . 21 : 2 = 1071
*\(\frac{x}{200}\)=\(\frac{1^2}{1.2}\).\(\frac{2^2}{2.3}\)....\(\frac{99^2}{99.100}\)
=>\(\frac{x}{200}\)=\(\frac{1}{2}\).\(\frac{2}{3}\)....\(\frac{99}{100}\)
=>\(\frac{x}{200}\)=\(\frac{1}{100}\)
=>100x=200
=>x=2
Đáp án cần chọn là: B
M = 1 2 1.2 . 2 2 2.3 . 3 2 3.4 ..... 99 2 99.100 . 100 2 101 = 1 2 .2 2 .3 2 ....99 2 .100 2 ( 1.2 ) . ( 2.3 ) . ( 3.4 ) .... ( 99.100 ) . ( 100.101 ) = ( 1.2.3.....99.100 ) . ( 1.2.3....99.100 ) ( 1.2.3.....99.100 ) . ( 2.3....99.100 ) .101 = 1 101