A=3/2.4+3/4.6+3/6.8+...3/49.51
B=5/1.3+5/3.5+3/5.7+...5/49.51
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a) =1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/101
=1-1/101
=100/101
b) =(2/1.3+2/3.5+2/5.7+...+2/99.101).2,5
=(1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/101).2,5
=(1-1/101).2,5
=100/101.2,5
=250/101
c) =(2/2.4+2/4.6+2/6.8+...+2/2008-2/2010).2
=(1/2-1/4+1/4-1/6+1/6-1/8+...+1/2008-1/2010).2
=(1/2-1/2010).2
=1004/1005
\(A=\) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(A=1-\frac{1}{50}\)
\(A=\frac{49}{50}\)
\(A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.50}\)
A= \(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\)
A = \(\frac{1}{1}-\frac{1}{51}=\frac{50}{51}\)
A = 13/21.2/11 + 13/21.9/11 + 8/21
= (13/21) + (13/21) + (8/21)
= (13 + 13 + 8)/21
= 34/21
B = (1 - 1/5)(1 - 2/5)(1 - 3/5)...(1 - 9/5)
= (4/5)(3/5)(2/5)(1/5)(0/5)(-1/5)(-2/5)(-3/5)(-4/5)
= 0
C = (1 - 1/2)(1 - 1/3)(1 - 1/4)...(1 - 1/50)
= (1/2)(2/3)(3/4)(4/5)...(49/50)
= 1/50
D = (2^2/1.3) * (3^2/2.4) * (4^2/3.5) * (5^2/4.6) * (6^2/5.7)
= (4/3) * (9/8) * (16/15) * (25/23) * (36/35)
= 0.979
Bài 1 :
\(S=1.3+3.5+5.7+...+99.101=3+15+35+...9999\)
Ta thấy :
\(3=2^2-1\)
\(15=4^2-1\)
\(35=6^2-1\)
.....
\(9999=100^2-1\)
\(\Rightarrow S=2^2+4^2+...+100^2-\left(1\right).\left(\left(100-2\right):2+1\right)\)
\(\Rightarrow S=\dfrac{100.\left(100+1\right)\left(2.100+1\right)}{6}-51\)
\(\Rightarrow S=\dfrac{100.101.201}{6}-51=338299\)
b: 6B=2*4*6+4*6*6+6*8*6+...+46*48*6+48*50*6
=2*4*6-2*4*6+4*6*8-4*6*8+...-44*46*48+46*48*50-46*48*50+48*50*52
=48*50*52
=>B=20800
d: 9D=1*4*9+4*7*9+...+46*49*9
=1*4*2+1*4*7-1*4*7+1*7*10-1*7*10+...+46*49*52-46*49*43
=1*2*4+46*49*52
=117216
=>D=13024
a:
\(A=\frac{3}{2\cdot4}+\frac{3}{4\cdot6}+...+\frac{3}{48\cdot50}\)---> Mik nghĩ bn ghi nhầm :]
\(A=\frac{3}{2}\left[\frac{1}{2\cdot4}+\frac{1}{4\cdot6}+...+\frac{1}{48\cdot50}\right]\)
\(A=\frac{3}{2}\left[\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{48}-\frac{1}{50}\right]\)
\(A=\frac{3}{2}\left[\frac{1}{2}-\frac{1}{50}\right]=\frac{3}{2}\cdot\frac{12}{25}=\frac{18}{25}\)
Vậy A = 18/25
\(B=\frac{5}{1\cdot3}+\frac{5}{3\cdot5}+...+\frac{5}{49\cdot51}\)
\(B=\frac{5}{2}\left[\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+...+\frac{1}{49\cdot51}\right]\)
\(B=\frac{5}{2}\left[\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{51}\right]\)
\(B=\frac{5}{2}\left[1-\frac{1}{51}\right]=\frac{5}{2}\cdot\frac{50}{51}=\frac{125}{51}\)
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