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\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{50-49}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}=\frac{49}{50}\)
\(B=1.2+2.3+3.4+...+49.50\)
\(3B=1.2.3+2.3.3+3.4.3+...+49.50.3\)
\(=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+49.50.\left(51-48\right)\)
\(=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+49.50.51-48.49.50\)
\(=49.50.51\)
\(B=\frac{49.50.51}{3}=49.50.17\)
\(50^2.A-\frac{B}{17}=49.50-49.50=0\)
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
=> 3A = 1.2.3 + 2.3.3 + 3.4.3 + .... + 49.50.3
=> 3A = 1.2.3 + 2.3.( 4 - 1 ) + 3.4.( 5 - 2 ) + .... + 49.50.( 51 - 48 )
=> 3A = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + .... + 49.50.51 - 48.49.50
=> 3A = ( 1.2.3 - 1.2.3 ) + ( 2.3.4 - 2.3.4 ) + .... + ( 48.49.50 - 48.49.50 ) + 49.50.51
=> 3A = 49.50.51
=> A = ( 49.50.51 ) : 3
=> A = 41650
A = 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+2.3.(4-1)+3.4.(5-2)+...+48.49.(50-47)+49.50.(51-48)
3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+48.49.50-47.48.49+49.50.51-48.49.50
3A=(1.2.3-1.2.3)+(2.3.4-2.3.4)+...(47.48.49-47.48.49)-(48.49.50-48.49.50)+49.50.51
3A=0+0+...+0+0+49.50.51
3A=49.50.51
A=\(\frac{49.50.51}{3}\)
A=41650
Đáp số: A=41650
Ta có : S = 1.2 + 2.3 + 3.4 + ..... + 32.33
=> 3S = 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + ...... + 32.33.34
=> 3S = 32.33.34
=> S = \(\frac{32.33.34}{3}=11968\)
A = 1.2 + 2.3 + ........+49.50
3A = 1.2.(3-0) + 2.3.(4-1)+........+49.50.(51 - 48)
3A = 1.2.3 + 2.3.4 - 1.2.3 +........ + 49.50.51 - 48.49.50
3A = 48.49.50 = 117600
A = 39200
Ta có :
Gọi A=1.2+2.3+3.4+4.5+...+49.50
A=1.2+2.3+3.4+4.5+...+49.50
3.A=3.(1.2+2.3+3.4+4.5+...+49.50)
3.A=1.2.3+2.3.3+3.3.4+3.4.5+...+3.49.50
3.A=1.2.(3-0)+2.3.(3-0)+(3-0).3.4+(3-0).4.5+...+(3-0).49.50
3.A=1.2.3-0+2.3.3-0+3.3.4-0+3.4.5-0+...+3.49.50-0
3.A=1.2.3-0+2.3.4-1.2.3+5.3.4-2.3.4+...+49.50.51-48.49.50
3.A=49.50.51
A= 49.50.51/3
A= 49.50.17.3/3
A=49.50.17
A=41650
Đáp số : A=41650
\(\dfrac{N}{2}=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{49\cdot50}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\\ =1-\dfrac{1}{50}< 1\\ N< 2\)
