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+Câu a:
A = 1/1.2 + 1/2.3 + ...+ 1/5.6 + 1
A = 1/1 - 1/2 + 1/2 - 1/3 + ...+ 1/5 - 1/6 + 1
A = 1/1 - 1/6 + 1
A = 6/6 - 1/6 + 6/6
A = 5/6 + 6/6
A = 11/6
Câu b:
B = 1/1.2.3 + 1/2.3.4 + 1/3.4.5 + ...+ 1/98.99.100
B = 1/2. (2/1.2.3 + 2/2.3.4 + ...+ 2/98.99.100)
B = 1/2.(1/1.2 - 1/2.3 + 1/2.3 - 1/3.4 + ...+ 1/98.99 - 1/99.100)
B = 1/2.(1/2 - 1/9900)
B = 1/2.4949/9900
B = 4949/19800
Câu a:
1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 + 1/5.6 + 1
= 1/1 - 1/2 + 1/2 - 1/3 + 1/4 - 1/5 + 1/5 - 1/6 + 1
= 1/1 - 1/6 + 1
= 6/6 - 1/6 + 6/6
= 5/6 + 1
= 11/6
Câu b:
Câu b:
B = 1/1.2.3 + 1/2.3.4 + 1/3.4.5 + ...+ 1/98.99.100
B = 1/2. (2/1.2.3 + 2/2.3.4 + ...+ 2/98.99.100)
B = 1/2.(1/1.2 - 1/2.3 + 1/2.3 - 1/3.4 + ...+ 1/98.99 - 1/99.100)
B = 1/2.(1/2 - 1/9900)
B = 1/2.4949/9900
B = 4949/19800
Câu a:
A = 1.2 + 2.3 + 3.4 + ...+ 99.100
3A = 1.2.3 + 2.3.3 +..+ 99.100.3
1.2.3 = 1.2.3
2.3.3 = 2.3(4 - 1) = 2.3.4 - 1.2.3
3.4.4 = 3.4(5 - 2) = 3.4.5 - 2.3.4
.............................................................
99.100.3 = 99.100.(101 - 98)=99.100.101-98.99.100
Cộng vế với vế ta có:
3A = 99.100.101
A = 99.100.101 : 3
A = 333300
Câu b:
1.3 + 3.5 + 5.7 + ..+ 97.99
6.A = 1.3.6 + 3.5.6 + 5.7.6 + ...+ 97.99.6
1.3.6 = 1.3.(5 + 1) = 1.3.5 + 1.3.1
3.5.6 = 3.5.(7 - 1) = 3.5.7 - 1.3.5
5.7.6 = 5.7.(9 - 3) = 5.7.9 - 3.5.7
.......................................................
97.99.6 = 97.99.(101 - 95) = 97.99.101-95.97.99
Cộng vế với vế ta có:
6A = 1.3.1 + 97.99.101
A = (1.3.1 + 97.99.101) : 6
A = (3 + 9603.101) : 6
A = (3+ 969903) : 6
A = 969906 : 6
A = 161651
A=1x2+2x3+3x4+...+49x50
3A= 3(1.2+2.3+3.4+...+49.50)
3A= 1.2.3+2.3.3+3.4.3+...+49.50.3
3A= 1.2.(3-0)+2.3(4-1)+3.4(5-2)+...+49.50.(51-48)
3A= 0.1.2-1.2.3+1.2.3-2.3.4+2.3.4-3.4.5+...+48.49.50-49.50.51
3A= 49.50.51
A= 49.50.51/3=41650
B=1x3+3x5+5x7+...+99x101
B=1/1.3 +1/3.5 +...+1/99.101
2B=2/1.3 + 2/3.5 +...+2/99.101
2B=1-1/3+1/3-1/5+...+1/99-1/101
2B=1-1/101
2B=100/101
B=100/101:2=100/202
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{x\left(x+1\right)}=\frac{99}{100}\)
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{x}-\frac{1}{x+1}=\frac{99}{100}\)
\(1-\frac{1}{x+1}=\frac{99}{100}\)
=> \(\frac{1}{x+1}=1-\frac{99}{100}=\frac{1}{100}\)
=> x+1 = 100
=> x = 100 - 1
=> x = 99
A = 1x2 + 2x3 + 3x4 + 4x5 + ...+ 99x100
A x 3 = 1x2x3 + 2x3x3 + 3x4x3 + 4x5x3 + ... + 99x100x3
A x 3 = 1x2x3 + 2x3x(4-1) + 3x4x(5-2) + 4x5x(6-3) + ... + 99x100x(101-98)
A x 3 = 1x2x3 + 2x3x4 - 1x2x3 + 3x4x5 - 2x3x4 + 4x5x6 - 3x4x5 + ... + 99x100x101 - 98x99x100.
A x 3 = 99x100x101
A = 99x100x101 : 3
A = 333300
Ta có:
\(A=1.2+2.3+3.4+...+99.100\)
\(\Rightarrow3A=1.2.\left(3-0\right)+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+99.100.\left(101-98\right)\)
\(\Rightarrow3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+99.100.101-98.99.100\)
\(\Leftrightarrow3A=99.100.101\Leftrightarrow A=\frac{99.100.101}{3}=333300\)
\(B=1.2.3+2.3.4+4.5.6+...+98.99.100\)
\(\Rightarrow4B=1.2.3.\left(4-0\right)+2.3.4.\left(5-1\right)+4.5.6.\left(7-3\right)+...+98.99.100.\left(101-97\right)\)
\(\Rightarrow4B=1.2.3.4+2.3.4.5-1.2.3.4+4.5.6.7-3.4.5.6+...+98.99.100.101-97.98.99.100\)
\(\Leftrightarrow4B=98.99.100.101\Leftrightarrow B=\frac{98.99.100.101}{4}=24497550\)
A= 1 x 2 + 2 x 3 + 3 x 4 +........+ 99 x 100
=> 2 + 6 + 12 +........+ 9900
=> 8 + 12 +.....+ 9900
=> 20 +....+ 9900
=> 20 + 20 + 30 +....+ 9900
=> 70 +....+ 9900
=> ( 9900 x 70 ) : 2
=> 693000 : 2
=> 346500
B = 1 x 2 x 3 + 2 x 3 x 4 +......+ 98 x 99 x100
=> ( 1 x 2 x 3) + ( 2 x 3 x 4 ) +....+ ( 98 x 99 x 100 )
= 6 + 24 +.......+ 970200
=> 28 + 120 +...+ 970200
=> ( 148 x 970200 ) : 2
=> 143589600 : 2
=> 71794900
A = 1.2 + 2.3 + .... + 99.100
3A = 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + ..... + 99.100.101
3A = 99.100.101
A = 99.100.101/3 = 333300
Ta có:
\(A=1\times2+2\times3+...+99\times100\)
\(3A=1\times2\times\left(3-0\right)+2\times3\times\left(4-1\right)+...+99\times100\times\left(101-98\right)\)
\(3A=1\times2\times3+2\times3\times4-1\times2\times3+...+99\times100\)
\(3A=99\times100\times101\)
\(A=\frac{99\times100\times101}{3}=333300\)
Vậy \(A=333300\)
Ta có:
\(B=1\times2\times3+2\times3\times4+...+98\times99\times100\)
\(4B=1\times2\times3\times\left(4-0\right)+2\times3\times4\times\left(5-1\right)+...+98\times99\times100\)
\(4B=1\times2\times3\times4+2\times3\times4\times5-1\times2\times3\times4+...+98\times99\times100\times101\)\(-97\times98\times99\times100\)
\(4B=98\times99\times100\times101\)
\(B=\frac{98\times99\times100\times101}{4}=24497550\)
Vậy \(B=24497550\)
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