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Đặt A=1.2.3.4+2.3.4.5+...+97.98.99.100
4A=(1.2.3+2.3.4+3.4.5+4.5.6+...+98.99.100)4
4A=1.2.3(4-0)+2.3.4(5-1)+3.4.5(6-2)+4.5.6(7-3)+...+98.99.100(101-97)
4A=1.2.3.4+2.3.4.5-1.2.3.4+3.4.5.6-2.3.4.5+4.5.6.7-3.4.5.6+...+98.99.100.101-97.98.99.100
4A=1.2.3.4-1.2.3.4+2.3.4.5-2.3.4.5+3.4.5.6-3.4.5.6+...+97.98.99.100-97.98.99.100+98.99.100.101
4A=98.99.100.101
A=98.99.100.101/4
5A=(5-0).1.2.3.4+(6-1).2.3.4.5+...+(101-96).97.98.99.100
5A=1.2.3.4.5-0+2.3.4.5.6-1.2.3.4.5+...+97.98.99.100.101-96.97.98.99.100
5A=97.98.99.100.101=9505049400
A=1901009880
Ta có \(\dfrac{1}{n\left(n+1\right)\left(n+2\right)}-\dfrac{1}{\left(n+1\right)\left(n+2\right)\left(n+3\right)}=\dfrac{3}{n\left(n+1\right)\left(n+2\right)\left(n+3\right)}\)
Áp dụng:
\(\dfrac{1}{1\cdot2\cdot3\cdot4}+\dfrac{1}{2\cdot3\cdot4\cdot5}+...+\dfrac{1}{27\cdot28\cdot29\cdot30}\\ =\dfrac{1}{3}\left(\dfrac{3}{1\cdot2\cdot3\cdot4}+\dfrac{3}{2\cdot3\cdot4\cdot5}+...+\dfrac{3}{27\cdot28\cdot29\cdot30}\right)\\ =\dfrac{1}{3}\left(\dfrac{1}{1\cdot2\cdot3}-\dfrac{1}{2\cdot3\cdot4}+\dfrac{1}{2\cdot3\cdot4}-\dfrac{1}{3\cdot4\cdot5}+...+\dfrac{1}{27\cdot28\cdot29}-\dfrac{1}{28\cdot29\cdot30}\right)\\ =\dfrac{1}{3}\left(\dfrac{1}{1\cdot2\cdot3}-\dfrac{1}{28\cdot29\cdot30}\right)\\ =\dfrac{1}{3}\left(\dfrac{1}{6}-\dfrac{1}{24360}\right)=\dfrac{1}{3}\cdot\dfrac{1353}{8120}=\dfrac{451}{8120}\)
\(\dfrac{1}{1.2.3.4}+\dfrac{1}{2.3.4.5}+\dfrac{1}{3.4.5.6}+...+\dfrac{1}{27.28.29.30}\)
\(=\dfrac{1}{3}\left(\dfrac{3}{1.2.3.4}+\dfrac{3}{2.3.4.5}+\dfrac{3}{3.4.5.6}+...+\dfrac{3}{27.28.29.30}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{1.2.3}-\dfrac{1}{2.3.4}+\dfrac{1}{2.3.4}-\dfrac{1}{3.4.5}+...+\dfrac{1}{27.28.29}-\dfrac{1}{28.29.30}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{1.2.3}-\dfrac{1}{28.29.30}\right)=\dfrac{1}{3}.\dfrac{4060-1}{28.29.30}\)
\(=\dfrac{1}{3}.\dfrac{4059}{24360}=\dfrac{1353}{24360}=\dfrac{451}{8120}\)
Lại phải giải hết
Gọi dãy số trên là A
\(A=\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+.....+\frac{1}{200.201.202.203}\)
\(3A=\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-.....+\frac{1}{200.201.202}-\frac{1}{201.202.203}\)
\(3A=\frac{1}{1.2.3}-\frac{1}{201.202.203}\)(chỗ này lm hơi tắt tí )
\(3A=\frac{1}{6}-\frac{1}{8242206}=\frac{1373701}{8242206}-\frac{1}{8242206}=\frac{1373700}{8242206}\)
\(A=\frac{1373700}{8242206}:3=\frac{457900}{8242206}\)
Câu 1:
A=1.2.3.4+2.3.4.5+3.4.5.6+...+2915.2916.2917.2918
5A=1.2.3.4.5+2.3.4.5.(6-1)+3.4.5.6(7-2)...+2915.2916.2917.2918(2919-2914)
5A=1.2.3.4.5+2.3.4.5.6-1.2.3.4.5+3.4.5.6.7-2.3.4.5.6.+...+2915.2916.2917.2918.2919-2914.2915.2916.2917.2918
5A=2915.2916.2917.2918.2919
A=2915.2916.2917.2918.2919/5
Câu 2:
Đáp án là: 541294159423242052710000000000000000000 (bấm máy tính chưa chắc đã đúng đâu)
Câu 3:
\(C=10,1+\frac{1993}{999900}=\frac{10100983}{999900}\)
Câu 4:
Chưa rõ phần vị trí mod mod j j đó
\(\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+...+\frac{1}{27.28.29.30}\)
\(=\frac{1}{3}\left(\frac{3}{1.2.3.4}+\frac{3}{2.3.4.5}+...+\frac{3}{27.28.29.30}\right)\)
\(=\frac{1}{3}\left(\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+...+\frac{1}{27.28.29}-\frac{1}{28.29.30}\right)\)
\(=\frac{1}{3}\left(\frac{1}{1.2.3}-\frac{1}{28.29.30}\right)\)
Ta có P có giá trị dương=> P>0
=> (2x-1)và(5-2x) cùng dấu âm hoặc dương
Xét (2x-1)>0=>x>\(\dfrac{1}{2}\)(1)
(5-2x)>0=>x<\(\dfrac{5}{2}\) (2)
Từ (1) và (2) =>x=1 hoặc x=2
Xét (2x-1)<0=>x<\(\dfrac{1}{2}\)(3)
(5x-2)<0=>x>\(\dfrac{5}{2}\)(4)
Từ (3) và (4) => x ko có giá trị nào
Vậy x=1 hoặc x=2
Đặt \(A=\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+...+\frac{1}{17.18.19.20}\)
\(A=\frac{4-1}{1.2.3.4}+\frac{5-2}{2.3.4.5}+....+\frac{20-17}{17.18.19.20}\)
\(A=\frac{3}{1.2.3.4}+\frac{3}{2.3.4.5}+....+\frac{3}{17.18.19.20}\)
\(3A=\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+....+\frac{1}{17.18.19}-\frac{1}{18.19.20}\)
\(3A=\frac{1}{1.2.3}-\frac{1}{18.19.20}=\frac{1139}{6840}\)
\(\Rightarrow A=\frac{1139}{6840}\div3=\frac{1139}{20520}\)
5P=(5-0).1.2.3.4+(6-1).2.3.4.5+...+(101-96).97.98.99.100
5P=1.2.3.4.5-0+2.3.4.5.6-1.2.3.4.5+....+97.98.99.100.101-96.97.98.99.100
5P=97.98.99.100.101
5P=9505049400
S=1901009880
P = 1.2.3.4 + 2.3.4.5 + 3.4.5.6 + 4.5.6.7 + .. + 97.98.99.100
4P = ( 1.2.3 + 2.3.4 + 3.4.5 + 4.5.6 + .. + 98.99.100) 4
4P = 1.2.3.(4-0) + 2.3.4(5-1) + 3.4.5(6-2) + 4.5.6(7-3) + 98.99.100(101-97)
4P = 1.2.3.4 + 2.3.4.5 - 1.2.3.4 + 3.4.5.6 - 2.3.4.5 + 4.5.6.7 - 3.4.5.6 + .. 98.99.100.101 - 97.98.99.100
4P = 98.99.100.101
4P= 98.99.100.101/4
Nếu thấy đúng thì tích mk nha