Tính tổng S = 3x4 + 4x5 + .......... + 49x50
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A
phân tích :
= 2 + 6 + 12 + 20 + 30 ... + 2450
quy luật : 2 số liền nhau hơn kém nhau là các số chẵn liên tiếp :
6 - 2 = 4 ; 12 - 6 = 6 ; 20 - 12 = 8
và bây giờ dùng tính chất dãy số để tính
nhé !
A×3=1.2.3+2.3.3+3.4.3+.......+49.50.3
A×3=1.2.(3-0)+2.3.(4-1)+3.4.(5-2)+.......+49.50.(51-48)
A×3=1.2.3-1.2.0+2.3.4-2.3.1+........+49.50.51-49.50.48
Ta thấy ngoài số 49.50.51 thì các số còn lại đều bị giản ước như 1.2.3 với 2.3.1;....nên
A×3=49.50.51
A×3=124950
A=124950:3
A=41650.
Vậy A=41650.
Ta có :
Gọi A=1.2+2.3+3.4+4.5+...+49.50
\Rightarrow A=1.2+2.3+3.4+4.5+...+49.50
\Rightarrow 3.A=3.(1.2+2.3+3.4+4.5+...+49.50)
\Rightarrow 3.A=1.2.3+2.3.3+3.3.4+3.4.5+...+3.49.50
\Rightarrow 3.A=1.2.(3-0)+2.3.(3-0)+(3-0).3.4+(3-0).4.5+...+(3-0).49.50
\Rightarrow 3.A=1.2.3-0+2.3.3-0+3.3.4-0+3.4.5-0+...+3.49.50-0
\Rightarrow 3.A=1.2.3-0+2.3.4-1.2.3+5.3.4-2.3.4+...+49.50.51-48.49.50
\Rightarrow 3.A=49.50.51
\Rightarrow A=49.50.51:3
\Rightarrow A=49.50.17.3:3
\Rightarrow A=49.50.17
\Rightarrow A=41650
Đáp số : A=41650
\(\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{49.50}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}\)
\(=\frac{1}{2}+\left(\frac{1}{3}-\frac{1}{3}\right)+\left(\frac{1}{4}-\frac{1}{4}\right)+\left(\frac{1}{5}-\frac{1}{5}\right)+...+\left(\frac{1}{49}-\frac{1}{49}\right)-\frac{1}{50}\)
\(=\frac{1}{2}-\frac{1}{50}=\frac{12}{25}\)
~ Hok tốt ~
\(\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{49.50}\)
\(=2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{50}\right)=2.\frac{12}{25}=\frac{24}{25}\)
Ta có:
\(\frac{1}{3.4}.x+\frac{1}{4.5}.x+...+\frac{1}{49.50}.x=1\)
\(\Rightarrow x.\left(\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{49.50}\right)=1\)
\(\Rightarrow x.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}\right)=1\)
\(\Rightarrow x.\left(\frac{1}{3}-\frac{1}{50}\right)=1\Leftrightarrow x.\frac{47}{150}=1\)
\(\Rightarrow x=1:\frac{47}{150}\Leftrightarrow x=\frac{150}{47}\)
S= 1x2+2x3+3x4+4x5+...+ 20x21
3xS=3x( 1x2+2x3+3x4+4x5+...+ 20x21 )
3xS = 1x2x3+2x3x3+3x4x3+....+20x21x3
3xS = 1x2x3 + 2x3x(4-1) + 3x4x(5-2)+........+20x21x(22-19)
3xS= 1x2x3 + 2x3x4 - 1x2x3 + 3x4x5 - 2x3x4 +......+20x21x22 - 19x20x21
3xS = 20x21x22
S = 20x21x22 /3
S= 1x2+2x3+3x4+4x5+...+ 20x21
3xS=3x( 1x2+2x3+3x4+4x5+...+ 20x21 )
3xS = 1x2x3+2x3x3+3x4x3+....+20x21x3
3xS = 1x2x3 + 2x3x(4-1) + 3x4x(5-2)+........+20x21x(22-19)
3xS= 1x2x3 + 2x3x4 - 1x2x3 + 3x4x5 - 2x3x4 +......+20x21x22 - 19x20x21
3xS = 20x21x22
S = 20x21x22 /3
k mk nha
c)
\(C=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{19.21}\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{19}-\frac{1}{21}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{21}\right)\)
\(=\frac{1}{2}.\frac{20}{21}\)
\(=\frac{10}{21}\)
\(A\)= \(\frac{1}{3.4}+\frac{1}{4.5}+..+\frac{1}{49.50}=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}=\)\(\frac{1}{3}-\frac{1}{50}=\frac{50}{150}-\frac{3}{150}=\frac{47}{150}\)
3S = 3 X4 X 3 + 4 X 5 X 3 +................+ 49 X 50 X 3
3S = 3 X 4 X ( 5 - 2 ) + 4 X 5 X ( 6 - 3 ) + .................. + 49 X 50 X ( 51 - 48 )
3S = 3 X 4 X 5 - 2 X 3 X 4 + 4 X 5 X 6 - 3 X 4 X 5 +............ + 49 X 50 X 51 - 48 X 49 X 50
3S = - ( 2 X 3 X 4 ) + ( 3 X 4 X 5 - 3 X 4 X 5 ) + ( 4 X 5 X 6 - 4 X 5 X 6 ) + ........... + ( 48 X 49 X 50 - 48 X 49 X 50 ) + 49 X 50 X 51
3S = - ( 2 X 3 X 4 ) + 0 + 0 + ...........+ 0 + 49 X 50 X 51
3S = - 24 + 124950
3S = 124926=> S = 124926 : 3 => S = 41642
Ta có : S = 3.4 + 4.5 + 5.6 +.....+49.50
=>3S = - 2.3.4 + 3.4.5 - 3.4.5 + ..... + 49.50.51
=> 3S = -24 + 49.50.51
=> 3S = -24 + 124950 = 124926
=> S = 124926/3 = 41642