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\(M=\dfrac{3}{2.4}+\dfrac{3}{4.6}+\dfrac{3}{6.8}+...+\dfrac{3}{100.102}\)
=> \(2M=\dfrac{3.2}{2.4}+\dfrac{3.2}{4.6}+\dfrac{3.2}{6.8}+...+\dfrac{3.2}{100.102}\)
=> \(2M=3.\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+...+\dfrac{2}{100.102}\right)\)
=> \(2M=3.\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{100}-\dfrac{1}{102}\right)\)
=> \(2M=3.\left(\dfrac{1}{2}-\dfrac{1}{102}\right)\)
=> \(2M=3.\dfrac{25}{51}\)
=> \(2M=\dfrac{25}{17}\)
=> \(M=\dfrac{25}{17}:2\)
=> \(M=\dfrac{25}{34}\)
A=(1.2)(2.2)+(2.2)(3.2)+...+(50.2)(51.2)
A=1.2.4+2.3.4+...+50.51.4
A=4(1.2+2.3+...+50.51)
M= 1.2+2.3+...+50.51
3M=1.2.3+2.3.(4-1)+...+50.51.(52-49)
=1.2.3+2.3.4-1.2.3+...+50.51.52-49.50.51
= 50.51.52
=132600
=> M=44200
=> A=4M=176800
\(B=2.4+4.6+6.8+...+98.100\)
\(B=2.\left(1.2\right)+2.\left(2.3\right)+2.\left(3.4\right)+...+2.\left(49.50\right)\)
\(B=2\left(1.2+2.3+3.4+....+49.50\right)\)
Đặt:
\(A=1.2+2.3+3.4+...+49.50\)
\(3A=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+49.50.\left(51-48\right)\)
\(3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+49.50.51-48.49.50\)
\(3A=49.50.51\)
\(A=\dfrac{49.50.51}{3}=41650\)
\(B=2A=41650.2=83300\)
3/2.(1/2-1/4+1/4-1/6+...+1/100-1/102)
3/2.(1/2-1/102)
3/2.25/51=25/34
A=1/2.4+1/4.6+........+1/100.102
A=1/2-1/4+1/4-1/6+.......+1/100-1/102
A=1/2-1/102
A=51/102-1/102
A=50/102
A=25/51
2
\(S1=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{100.102}\)
\(S1=\frac{1}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{100.102}\right)\)
\(S1=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{100}-\frac{1}{102}\right)\)
\(S1=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{102}\right)\)
\(S1=\frac{1}{2}.\left(\frac{51}{102}-\frac{1}{102}\right)\)
\(S1=\frac{1}{2}.\frac{25}{51}\)
\(S1=\frac{25}{102}\)
\(A = 1.4 + 2.5 + 3.6 + ...+ 99.102\)
\(A=1.2+1.2+2.3+2.2+3.4+3.2+...+99.100+99.2\)
\(A=(1.2+2.3+3.4+...+99.100)+2.(1+2+3+...+99)\)
\(A=333300+9900\)
\(A=343200\)
\(B = 2.4 + 4.6 + 6.8 + ....+ 98.100 + 100.102\)
\(B=(1.2)(2.2)+(2.2)(3.2)+...+(50.2)(51.2) \)
\(B=4(1.2+2.3+...+50.51) \)
\(M= 1.2+2.3+...+50.51 \)
\(3M=1.2.3+2.3.(4-1)+...+50.51.(52-49) \)
\(=1.2.3+2.3.4-1.2.3+...+50.51.52-49.50.51 \)
\(= 50.51.52\)
\(=132600 \)
\(\Rightarrow\)\(M=44200 \)
\(\Rightarrow\) \(B=4M=176800\)
Bài 1 :
\(S=1.3+3.5+5.7+...+99.101=3+15+35+...9999\)
Ta thấy :
\(3=2^2-1\)
\(15=4^2-1\)
\(35=6^2-1\)
.....
\(9999=100^2-1\)
\(\Rightarrow S=2^2+4^2+...+100^2-\left(1\right).\left(\left(100-2\right):2+1\right)\)
\(\Rightarrow S=\dfrac{100.\left(100+1\right)\left(2.100+1\right)}{6}-51\)
\(\Rightarrow S=\dfrac{100.101.201}{6}-51=338299\)
Đặt A = 8.10 + 10.12 + 12.14 + ....... + 98.100
=> 6A = 8.10.12 - 8.10.12 + 10.12.14 - 10.12.14 + ...... + 98.100.102
=> 6A = 98.100.102
=> A = 98.100.102/6
=> A = 166600
c.1.2.3+2.3.4+4.5.6+5.6.7=6+24+120+210
=30+120+210
=150+210
=360
\(M=\dfrac{3}{2.4}+\dfrac{3}{4.6}+\dfrac{3}{6.8}+.....+\dfrac{3}{100.102}\)
\(M=\dfrac{3.2}{2.4}+\dfrac{3.2}{4.6}+\dfrac{3.2}{6.8}+.....+\dfrac{3.2}{100.102}\)
\(M=3.(\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+......+\dfrac{2}{100.102})\)
\(M=3.(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+.....+\dfrac{1}{100}-\dfrac{1}{102})\)
\(M=3.(\dfrac{1}{2}-\dfrac{1}{102})\)
\(M=3.\dfrac{50}{102}\)
\(M=\dfrac{25}{17}\)
Nếu ai mong bn thông cảm!!! Chúc bn hc tốt!
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