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16 tháng 4 2019

\(A=1.2+2.3+3.4+...+n\left(n+1\right)\)

\(\Rightarrow3A=1.2.3+2.3.3+3.4.3+...+n\left(n+1\right).3\)

\(\Rightarrow3A=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+n\left(n+1\right).\)\(\left(n+2-n+1\right)\)

\(\Rightarrow3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+n\left(n+1\right)\left(n+2\right)\)\(-\left(n-1\right)n\left(n+1\right)\)

\(\Rightarrow3A=n\left(n+1\right)\left(n+2\right)\)

\(\Rightarrow A=\frac{n\left(n+1\right)\left(n+2\right)}{3}\)

Vì A là số tự nhiên nên A chia hết cho 3 (đpcm)

16 tháng 5 2017

Ta có: \(\frac{1}{1.2}=\frac{3}{1.2.3}\) ;\(\frac{1}{1.2+2.3}=\frac{3}{2.3.4}\)\(\frac{1}{2.3+3.4}=\frac{3}{3.4.5}\); ......;\(\frac{1}{1.2+2.3+3.4+...+n\left(n+1\right)}=\frac{3}{n\left(n+1\right)\left(n+2\right)}\)

=> \(S=\frac{3}{1.2.3}+\frac{3}{2.3.4}+\frac{3}{3.4.5}+...+\frac{3}{n\left(n+1\right)\left(n+2\right)}\)

=> \(\frac{2S}{3}=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{n\left(n+1\right)\left(n+2\right)}\)

Ta lại có: \(\frac{2}{1.2.3}=\frac{1}{1.2}-\frac{1}{2.3}\)\(\frac{2}{2.3.4}=\frac{1}{2.3}-\frac{1}{3.4}\)\(\frac{2}{3.4.5}=\frac{1}{3.4}-\frac{1}{4.5}\);....;\(\frac{2}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)

=> \(\frac{2S}{3}=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)

=> \(\frac{2S}{3}=\frac{1}{1.2}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)=> \(S=\frac{3}{4}-\frac{3}{2\left(n+1\right)\left(n+2\right)}< \frac{3}{4}\)

=> \(S< \frac{3}{4}\)

16 tháng 5 2017

Mình nhầm 1 chỗ: \(\frac{1}{1.2+2.3+3.4}=\frac{3}{3.4.5}\)

31 tháng 7 2015

A=1.2+2.3+...+n(n+1)

3A=1.2.3+2.3.3+....+3n(n+1)

3A=1.2.3-0.1.2+2.3.4-1.2.3+3.4.5-2.3.4+...+n(n+1)(n+2)-(n-1)n(n+1)

3A=n(n+1)(n+2)

A=n(n+1)(n+2)/3 (đpcm)

31 tháng 7 2015

A=1.2+2.3+....+n(n+1)

3A=1.2.3+2.3.3+....+3n(n+1)

3A=1.2.3-0.1.2+2.3.4-1.2.3+3.4.5-2.3.4+...+n(n+1)(n+2)-(n-1)n(n+1)

3A=n(n+1)(n+2)

A=n(n+1)(n+2)/3 (đpcm)

 

26 tháng 2 2017

\(3D_n=1.2.3+2.3.3+3.4.3+...+n\left(n+1\right)3\)

\(=1.2\left(3-0\right)+2.3\left(4-1\right)+...+n\left(n+1\right)\left(n+2-n+1\right)\)

\(=1.2.3-0.1.2+2.3.4-1.2.3+...+n\left(n+1\right)\left(n+2\right)-\left(n-1\right)n\left(n+1\right)\)

\(=n\left(n+1\right)\left(n+2\right)-0.1.2=n\left(n+1\right)\left(n+2\right)\)

\(\Rightarrow D_n=\frac{n\left(n+1\right)\left(n+2\right)}{3}\)

26 tháng 2 2017

Dn = 1.2 + 2.3 + 3.4 +...+ n(n + 1)

3Dn = 1.2.(3 - 0) + 2.3.(4 - 1) + 3.4.(5 - 2) +...+ n(n + 1).[(n + 2) - (n - 1)]

3Dn = 1.2.3 - 0.1.2 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 +...+ n(n + 1)(n + 2) - (n - 1)n(n + 1)

3Dn = [1.2.3 + 2.3.4 + 3.4.5 +...+ n(n + 1)(n + 2)] - [0.1.2 + 1.2.3 + 2.3.4 +...+ n(n - 1)(n + 1)]

3Dn = n(n + 1)(n + 2) - 0.1.2

3Dn = n(n + 1)(n + 2)

Dn = \(\frac{n\left(n+1\right)\left(n+2\right)}{3}\) (đpcm)

29 tháng 12 2017

Đặt \(A=1.2+2.3+3.4+...+n\left(n+1\right)\)

\(\Rightarrow3A=1.2.3+2.3.3+3.4.3+...+3n\left(n+1\right)\)

\(=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+n\left(n+1\right)\left(n+2-n+1\right)\)

\(=1.2.3+2.3.4-1.2.3+...+n\left(n+1\right)\left(n+2\right)-\left(n-1\right)n\left(n+1\right)\)

\(=n\left(n+1\right)\left(n+2\right)\)

\(\Rightarrow1.2+2.3+3.4+...+n\left(n+1\right)=\frac{n\left(n+1\right)\left(n+2\right)}{3}\)

29 tháng 12 2017

Bạn ơi tại sao 3n.(n+1) lại bằng với n.(n+1).(n+2-n+1)

13 tháng 1 2016

 

D = 1.2 + 2.3+ 3.4 +...+ 99.100

=>3D=1.2.3+2.3.3+3.4.3+...+99.100.3

=1.2.(3-0)+2.3.(4-1)+3.4.(5-2)+....+99.100.(101-98)

=1.2.3-0.1.2+2.3.4-1.2.3+3.4.5-2.3.4+...+99.100.101-98.99.100

=99.100.101-0.1.2

=99.100.101

=999900

=>D=999900:3=333300

 

Dn = 1.2 + 2.3 + 3.4 +...+ n (n +1)

=>3Dn=1.2.3+2.3.3+3.4.3+...+n(n+1).3

=1.2.(3-0)+2.3.(4-1)+3.4.(5-2)+...+n.(n+1).[(n+2)-(n-1)]

=1.2.3-0.1.2+2.3.4-1.2.3+2.3.4-2.3.4+....+n(n+1)(n+2)-(n-1)n(n+1)

=n.(n+1).(n+2)-0.1.2

=n.(n+1)(n+2)

=>Dn=n.(n+1)(n+2):3

 =>điều cần chứng minh

20 tháng 10 2020

\(A=1.2+2.3+3.4+.......+\left(n-1\right).n\)

\(\Rightarrow3A=1.2.3+2.3.3+3.4.3+......+\left(n-1\right).n.3\)

\(=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+.....+\left(n-1\right).n.\left[\left(n+1\right)-\left(n-2\right)\right]\)

\(=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+......+\left(n-1\right).n\left(n+1\right)-\left(n-1\right).n\left(n-2\right)\)

\(=\left(n-1\right).n.\left(n+1\right)\)

\(\Rightarrow A=\frac{\left(n-1\right).n.\left(n+1\right)}{3}\)( đpcm )

20 tháng 10 2020

lol why lol 

AH
Akai Haruma
Giáo viên
30 tháng 6

Lời giải:

$A=1.2+2.3+3.4+...+(n-1)n$

$3A=1.2(3-0)+2.3(4-1)+3.4(5-2)+....+(n-1)n[(n+1)-(n-2)]$

$=[1.2.3+2.3.4+3.4.5+...+(n-1)n(n+1)]-[1.2.3+2.3.4+....+(n-2)(n-1)n]$

$=(n-1)n(n+1)$

$\Rightarrow A=\frac{n(n-1)(n+1)}{3}$

23 tháng 4 2023

\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{n\left(n+1\right)}\)

\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{n}-\dfrac{1}{n+1}\)

= 1 - \(\dfrac{1}{n+1}\) = \(\dfrac{n}{n+1}\)