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5 tháng 6 2016

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 

 

5 tháng 6 2016

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

14 tháng 8 2016

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

14 tháng 8 2016

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

26 tháng 9 2021

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:

\(\dfrac{1}{1\cdot2\cdot3\cdot4}+\dfrac{1}{2\cdot3\cdot4\cdot5}+...+\dfrac{1}{27\cdot28\cdot29\cdot30}\\ =\dfrac{1}{3}\left(\dfrac{3}{1\cdot2\cdot3\cdot4}+\dfrac{3}{2\cdot3\cdot4\cdot5}+...+\dfrac{3}{27\cdot28\cdot29\cdot30}\right)\\ =\dfrac{1}{3}\left(\dfrac{1}{1\cdot2\cdot3}-\dfrac{1}{2\cdot3\cdot4}+\dfrac{1}{2\cdot3\cdot4}-\dfrac{1}{3\cdot4\cdot5}+...+\dfrac{1}{27\cdot28\cdot29}-\dfrac{1}{28\cdot29\cdot30}\right)\\ =\dfrac{1}{3}\left(\dfrac{1}{1\cdot2\cdot3}-\dfrac{1}{28\cdot29\cdot30}\right)\\ =\dfrac{1}{3}\left(\dfrac{1}{6}-\dfrac{1}{24360}\right)=\dfrac{1}{3}\cdot\dfrac{1353}{8120}=\dfrac{451}{8120}\)

 

26 tháng 9 2021

\(\dfrac{1}{1.2.3.4}+\dfrac{1}{2.3.4.5}+\dfrac{1}{3.4.5.6}+...+\dfrac{1}{27.28.29.30}\)

\(=\dfrac{1}{3}\left(\dfrac{3}{1.2.3.4}+\dfrac{3}{2.3.4.5}+\dfrac{3}{3.4.5.6}+...+\dfrac{3}{27.28.29.30}\right)\)

\(=\dfrac{1}{3}\left(\dfrac{1}{1.2.3}-\dfrac{1}{2.3.4}+\dfrac{1}{2.3.4}-\dfrac{1}{3.4.5}+...+\dfrac{1}{27.28.29}-\dfrac{1}{28.29.30}\right)\)

\(=\dfrac{1}{3}\left(\dfrac{1}{1.2.3}-\dfrac{1}{28.29.30}\right)=\dfrac{1}{3}.\dfrac{4060-1}{28.29.30}\)

\(=\dfrac{1}{3}.\dfrac{4059}{24360}=\dfrac{1353}{24360}=\dfrac{451}{8120}\)

14 tháng 1 2017

\(\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+...+\frac{1}{27.28.29.30}\)

\(=\frac{1}{3}\left(\frac{3}{1.2.3.4}+\frac{3}{2.3.4.5}+...+\frac{3}{27.28.29.30}\right)\)

\(=\frac{1}{3}\left(\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+...+\frac{1}{27.28.29}-\frac{1}{28.29.30}\right)\)

\(=\frac{1}{3}\left(\frac{1}{1.2.3}-\frac{1}{28.29.30}\right)\)

9 tháng 7 2017

Đặt \(A=\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+...+\frac{1}{17.18.19.20}\)

\(A=\frac{4-1}{1.2.3.4}+\frac{5-2}{2.3.4.5}+....+\frac{20-17}{17.18.19.20}\)

\(A=\frac{3}{1.2.3.4}+\frac{3}{2.3.4.5}+....+\frac{3}{17.18.19.20}\)

\(3A=\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+....+\frac{1}{17.18.19}-\frac{1}{18.19.20}\)

\(3A=\frac{1}{1.2.3}-\frac{1}{18.19.20}=\frac{1139}{6840}\)

\(\Rightarrow A=\frac{1139}{6840}\div3=\frac{1139}{20520}\)

AH
Akai Haruma
Giáo viên
12 tháng 7 2023

Lời giải:

$A=10(\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+...+\frac{1}{92.93.94.95})$
$3A=10(\frac{4-1}{1.2.3.4}+\frac{5-2}{2.3.4.5}+...+\frac{95-92}{92.93.94.95})$

$=10(\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+...+\frac{1}{92.93.94}-\frac{1}{93.94.95})$

$=10(\frac{1}{1.2.3}-\frac{1}{93.94.95})$

$A=\frac{10}{3}(\frac{1}{1.2.3}-\frac{1}{93.94.95})$

12 tháng 7 2023

\(A=\dfrac{10}{1.2.3.4}+\dfrac{10}{2.3.4.5}+...+\dfrac{10}{92.93.94.95}\)

\(A=10.\left(\dfrac{1}{1.2.3.4}+\dfrac{1}{2.3.4.5}+...+\dfrac{1}{92.93.94.95}\right)\)

\(3A=10.\left(\dfrac{3}{1.2.3.4}+\dfrac{3}{2.3.4.5}+...+\dfrac{3}{92.93.94.95}\right)\)

\(3A=10.\left(\dfrac{1}{1.2.3}-\dfrac{1}{2.3.4}+\dfrac{1}{2.3.4}+...+\dfrac{1}{92.93.94}-\dfrac{1}{93.94.95}\right)\)

\(3A=10.\left(\dfrac{1}{1.2.3}-\dfrac{1}{93.94.95}\right)\)

\(3A=10.\left(\dfrac{138415-1}{93.94.95}\right)=\dfrac{1384140}{93.94.95}\)

\(A=\dfrac{461380}{93.94.95}=\dfrac{46138}{93.47.19}=\dfrac{46138}{83049}\)

\(\)

8 tháng 10 2015

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 )
\(3A=\frac{1}{6}-\frac{1}{8242206}=\frac{1373701}{8242206}-\frac{1}{8242206}=\frac{1373700}{8242206}\)
\(A=\frac{1373700}{8242206}:3=\frac{457900}{8242206}\)