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a) \(A=\frac{1}{8}+\frac{1}{24}+\frac{1}{48}+...+\frac{1}{10200}\)
\(A=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{100.102}\)
\(2A=\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{100.102}\)
\(2A=\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{6}\right)+\left(\frac{1}{6}-\frac{1}{8}\right)+...+\left(\frac{1}{100}-\frac{1}{102}\right)\)
\(2A=\frac{1}{2}-\frac{1}{102}\)
\(2A=\frac{25}{51}\)
\(A=\frac{25}{51}:2\)
\(A=\frac{25}{102}\)
Vậy \(\frac{1}{8}+\frac{1}{24}+\frac{1}{48}+...+\frac{1}{10200}=\frac{25}{102}\)
b) \(B=\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+...+\frac{3}{2015.2016}\)
\(B=3.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2015.2016}\right)\)
\(B=3.\left[\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+...+\left(\frac{1}{2015}-\frac{1}{2016}\right)\right]\)
\(B=3.\left(\frac{1}{1}-\frac{1}{2016}\right)\)
\(B=3.\frac{2015}{2016}\)
\(B=\frac{2015}{672}\)
Vậy \(\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+...+\frac{3}{2015.2016}=\frac{2015}{672}\)
Tính tổng :
a ,1+2+3+..........+2015
SSH của tổng trên là :
(2015-1):1+1=2015(SH)
Tổng trên là:
(2015:2)x(2015+1)=2031120
b, 3+5+7+......+2015
SSH của tổng trên là :
(2015-3):2+1=1007(SH)
Tổng trên là:
(1007:2)x(2015+3)=1016063
LƯU ý: SSH=số số hạng nha
\(S=\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+\frac{3}{4.5}+....+\frac{3}{2015.2016}\)
\(\Rightarrow\frac{1}{3}.S=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{2015.2016}\)
\(\Rightarrow\frac{1}{3}.S=\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+......+\left(\frac{1}{2015}-\frac{1}{2016}\right)\)
\(\Rightarrow\frac{1}{3}.S=\frac{1}{1}-\frac{1}{2016}\)
\(\Rightarrow\frac{1}{3}.S=\frac{2015}{2016}\)
\(\Rightarrow S=\frac{2015}{672}\)
Vậy: \(\Rightarrow S=\frac{2015}{672}\)
Bạn giải giúp mk câu mk đăng tầm 5 phút nha!
A= 1.2+2.3+3.4+...+2015.2016
3A=1.2.3+2.3.3+3.4.3+...+2015.2016.3
3A=1.2.3+2.3.(4-1)+3.4.(5-2)+...+2015.2016.(2017-2014)
3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+2015.2016.2017-2014.2015.2016
3A=2015.2016.2017
3A=8193538080
A=8193538080:3
A=2731179360
3A = 1.2.3 + 2.3.3 + 3.4.3 + ..... + 2015.2016.3
=> 3A = 1.2.3 + 2.3.( 4 -1 ) + 3.4.( 5 - 2 ) + .... + 2015.2016.( 2017 - 2014 )
=> 3A = 1.2.3 + 2.3.4 - 1.2.3 + .... + 2015.2016.2017 - 2014.2015.2016
=> 3A = 2015.2016.2017
=> A = \(\frac{2015.2016.2017}{3}\)
A=1.2+2.3+3.4+...+2015.2016
=> 3A=1.2.3+2.3.3+3.4.3+...+2015.2016.3
=> 3A=1.2.3+2.3.(4-1)+3.4.(5-2)+...+2015.2016.(2017-2014)
=>3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+ 2015.2016.2017-2014.2015.2016
=> 3A=2015.2016.2017
=> A=\(\frac{2015.2016.2017}{3}=2731179360\)
.
.
Đặt \(A=1.2+2.3+3.4+...+2015.2016\)
\(\Rightarrow3A=1.2.3+2.3.3+3.4.3+...+2015.2016.3\)
\(\Rightarrow3A=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+2015.2016.\left(2017-2014\right)\)
\(\Rightarrow3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+2015.2016.2017-2014.2015.2016\)
\(\Rightarrow3A=2015.2016.2017\)
\(\Rightarrow A=2015.2016.2017:3\)
\(\Rightarrow A=2015.672.2017\)
Vậy \(A=2015.672.2017\)
1 . 2 + 2 . 3 + 3 . 4 + ... + 2015 . 2016
3M = 1 . 2 . 3 + 2 . 3 . 3 + 3 . 4 . 3 + ... + 2015 . 2016 . 3
3M = 1 . 2 ( 3 - 0 ) + 2 . 3 ( 4 - 1 ) + 3 . 4 ( 5 - 2 ) + ... + 2015 . 2016 ( 2017 - 2014 )
3M = ( 1 . 2 . 3 + 2 . 3 . 4 + 3 . 4. 5 + ... + 2015 . 2016 . 2017 ) - ( 0 . 1 . 2 + 1 . 2 . 3 + 2 . 3 . 4 + ... + 2014 . 2015 . 2016 )
3M = 2015 . 2016 . 2017
M = \(\frac{2015.2016.2017}{3}\)
M = 2731179360
tách tử thành 1.3 ( cho 3 ra ngoài làm nhân tử chung)
=> ở mẫu còn nguyên tắc số thứ 2- số thứ 1 = tử
=> (1/1.2+1/2.3+.......+1/2015.2016 ) .3
= (2-1/1.2+3-2/2.3+......+2016-2015/2015.2016).3
= (2/1.2-1/1.2+3/2.3-2/2.3..........+2016/2015.2016- 2015/2015.2016).3
= ( 1-1/2+1/2-1/3+...........+ 1/2015-1/2016).3
= ( 1-1/2016 ) .3
= 2015/2016 .3
\(S=3.\left(\frac{1}{1}.\frac{1}{2}+\frac{1}{2}.\frac{1}{3}+\frac{1}{3}.\frac{1}{4}+\frac{1}{4}.\frac{1}{5}+...+\frac{1}{2015}.\frac{1}{2016}\right)\)
\(3S=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2015}-\frac{1}{2016}\)
\(3S=1-\frac{1}{2016}\)
\(3S=\frac{2015}{2016}\)
\(S=\frac{2015}{2016}:3\)
\(S=\frac{2015}{6048}\)
Đặt A = 1.2 + 2.3 + 3.4 + ... +2015.2016
3A = 1.2.3 + 2.3.(4-1) + ... + 2015.2016.(2017-2014)
3A = 1.2.3 + 2.3.4 - 1.2.3 + ... + 2015.2016.2017 - 2014.2015.2016
3A = 2014.2015.2016
A = 2727117120
\(S=\dfrac{3}{1.2}+\dfrac{3}{2.3}+...+\dfrac{3}{2015.2016}\)
\(=3\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{2015.2016}\right)\)
\(=3\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2015}-\dfrac{1}{2016}\right)\)
\(=3\left(1-\dfrac{1}{2016}\right)\)
\(=3.\dfrac{2015}{2016}=\dfrac{6045}{2016}\)
Vậy \(S=\dfrac{6045}{2016}\)
\(S=3\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2015.2016}\right)\)
\(\Rightarrow S=3\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2015}-\dfrac{1}{2016}\right)\)
\(\Rightarrow S=3\left(1-\dfrac{1}{2016}\right)=3.\dfrac{2015}{2016}=\dfrac{6045}{2016}\)
Vậy ...
\(S=\dfrac{3}{1.2}+\dfrac{3}{2.3}+\dfrac{3}{3.4}+....+\dfrac{3}{2015.2016}\)
\(=\dfrac{1}{1}.\left(\dfrac{3}{1.2}+\dfrac{3}{2.3}+\dfrac{3}{3.4}+....+\dfrac{3}{2015.2016}\right)\)
= \(3.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+....\dfrac{1}{2015.2016}\right)\)
= 3. \(\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2015}+\dfrac{1}{2016}\right)\)
= 3.
sorry em mình nghịch nên vẫn thiếu
bổ sung nè thêm vào chỗ " 3." nhé
= 3.\(\left(\dfrac{1}{1}-\dfrac{1}{2016}\right)\)
= 3.\(\dfrac{2015}{2016}\)
= \(\dfrac{6045}{2016}\)
Đúng thì tick nhé cả câu trả lời lúc nẫy ghép lại đấy, xin lỗi vì em tớ phá nên nó bị cách làm hai câu trả lời hưng thực ra là 1
\(S=\dfrac{3}{1.2}+\dfrac{3}{2.3}+\dfrac{3}{3.4}+...+\dfrac{3}{2015.2016}\)
\(3S=3.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2015.2016}\right)\)
\(3S=\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2015}+\dfrac{1}{2016}\right)\)
\(3S=1-\dfrac{1}{2016}\)
\(S=3.\dfrac{2015}{2016}=\dfrac{6045}{2016}\)
Vậy.................................
thanks nhiều
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