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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)

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)

11 tháng 5 2020

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)

\(A=1-\frac{1}{6}=\frac{5}{6}\)

\(B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{n}-\frac{1}{n+1}\)

\(B=1-\frac{1}{n+1}=\frac{n}{n+1}\)

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)

\(A=-\frac{1}{1.2}-\frac{1}{2.3}-\frac{1}{3.4}-...-\frac{1}{\left(n-1\right)n}\)

\(=-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{n-1.n}\right)\)

\(=-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}\right)\)

\(=-\left(1-\frac{1}{n}\right)\)

\(=-\frac{n-1}{n}\)

16 tháng 9 2019

\(A=-\frac{1}{1.2}-\frac{1}{2.3}-\frac{1}{3.4}-...-\frac{1}{\left(n-1\right).n}\)

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

\(A=-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{\left(n-1\right)}-\frac{1}{n}\right)\)

\(\Rightarrow A=-\left(1-\frac{1}{n}\right)\)

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

2 tháng 5 2019

P=3 /1.2+1/22.32+...+4033/20162.20172

P=1/1 -1/22 +1/22 -1/52 +...+1/2016- 1/20172

P=1-1/20172 <1

vậy p<1

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)

22 tháng 4 2018

G= \(\frac{1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+98\right)}{1.2+2.3+3.4+...+98.99}\)

G= \(\frac{\frac{1.2}{2}+\frac{2.3}{2}+\frac{3.4}{2}+...+\frac{98.99}{2}}{1.2+2.3+3.4+...+98.99}\)

G = \(\frac{\frac{1.2+2.3+...+98.99}{2}}{1.2+2.3+3.4+...+98.99}\)

G= \(\frac{1}{2}\)

22 tháng 2 2017

mh chịu thôi