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\(\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+...+\dfrac{2}{98.100}\\ =\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{98}-\dfrac{1}{100}\\ =\dfrac{1}{2}-\dfrac{1}{100}\\ =\dfrac{49}{100}\)
mình k viết lại đề nhé =)
câu A
A :5 =1/2.4+1/4.6+1/6.8+..+1/98.100
A:5 =1/2-1/4+1/4-1/6+1/6-1/8+...+1/98-1/100
A:5 =1/2-1/100 =49/100
A=49/100 x5 =49/20
câu B tươg tự nha =)
Ta có:
A =5/2(1/2-1/4 + 1/4-1/6+ 1/6..........1/98-1/100)
A =5/2 (1/2 -1/100)
A =5/2 x 49/100
A = 49/20
a,6B=2.4.6+4.6.(8-2)+...............+98.100.(102-96)
6B=2.4.6+4.6.8-2.4.6+..............+98.100.102-96.98.100
6B=98.100.102
B=98.100.102:6
B=166600
\(B=2.4+4.6+6.8+...+98.100\)
\(B=2.\left(1.2\right)+2.\left(2.3\right)+2.\left(3.4\right)+...+2.\left(49.50\right)\)
\(B=2\left(1.2+2.3+3.4+....+49.50\right)\)
Đặt:
\(A=1.2+2.3+3.4+...+49.50\)
\(3A=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+49.50.\left(51-48\right)\)
\(3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+49.50.51-48.49.50\)
\(3A=49.50.51\)
\(A=\dfrac{49.50.51}{3}=41650\)
\(B=2A=41650.2=83300\)
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
M = \(\dfrac{5}{2.4}\) + \(\dfrac{5}{4.6}\)+ \(\dfrac{5}{6.8}\)+ ...+ \(\dfrac{5}{96.98}\)+ \(\dfrac{5}{98.100}\)
M = \(\dfrac{5}{2}\).( \(\dfrac{2}{2.4}\) + \(\dfrac{2}{4.6}\)+ \(\dfrac{2}{6.8}\)+...+ \(\dfrac{2}{96.98}\)+ \(\dfrac{2}{98.100}\))
M = \(\dfrac{5}{2}\).( \(\dfrac{1}{2}-\dfrac{1}{4}\)+ \(\dfrac{1}{4}-\dfrac{1}{6}\)+ \(\dfrac{1}{6}\) - \(\dfrac{1}{8}\)+...+ \(\dfrac{1}{96}\)-\(\dfrac{1}{98}\)+ \(\dfrac{1}{98}\)-\(\dfrac{1}{100}\))
M = \(\dfrac{5}{2}\).(\(\dfrac{1}{2}\) - \(\dfrac{1}{100}\))
M = \(\dfrac{49}{40}\)
\(x\) \(\times\) M - 1 = \(\dfrac{20}{29}\)
\(x\) \(\times\) \(\dfrac{49}{40}\) = \(\dfrac{20}{29}\) + 1
\(x\) \(\times\) \(\dfrac{49}{40}\) = \(\dfrac{49}{29}\)
\(x\) = \(\dfrac{49}{29}\) : \(\dfrac{49}{40}\)
\(x\) = \(\dfrac{40}{29}\)
\(A = 1.4 + 2.5 + 3.6 + ...+ 99.102\)
\(A=1.2+1.2+2.3+2.2+3.4+3.2+...+99.100+99.2\)
\(A=(1.2+2.3+3.4+...+99.100)+2.(1+2+3+...+99)\)
\(A=333300+9900\)
\(A=343200\)
\(B = 2.4 + 4.6 + 6.8 + ....+ 98.100 + 100.102\)
\(B=(1.2)(2.2)+(2.2)(3.2)+...+(50.2)(51.2) \)
\(B=4(1.2+2.3+...+50.51) \)
\(M= 1.2+2.3+...+50.51 \)
\(3M=1.2.3+2.3.(4-1)+...+50.51.(52-49) \)
\(=1.2.3+2.3.4-1.2.3+...+50.51.52-49.50.51 \)
\(= 50.51.52\)
\(=132600 \)
\(\Rightarrow\)\(M=44200 \)
\(\Rightarrow\) \(B=4M=176800\)
6A = 2 . 4 . 6 + 4.6.6 + 6.8.6 + ... + 98 . 100 . 6 = 2.4.6 + 4.6.(8-2) + ... + 98. 100 . (102 - 96)
= 2.4.6 +4.6.8 - 2.4.6 + .... + 98.100 . 102 - 96.98.100
= 98. 100 . 102
= 999600
Suy ra A = 166600
Vậy ______________________
\(\frac{2.4+4.6+6.8+...+98.100}{1.2+2.3+3.4+...+49.50}=\frac{4.\left(1.2+2.3+3.4+...+49.50\right)}{1.2+2.3+3.4+...+49.50}=\frac{4}{1}=4\)
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