Bài giải:

3S = 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 + 2.3.4 - 2.3 + 3.4.5 - 2.3.4 + ... + n(n + 1)(n + 2) - n(n + 1)(n - 1)

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

=> S N(N+1)(n+2)/3

3S = 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 + 2.3.4 - 2.3 + 3.4.5 - 2.3.4 + ... + n(n + 1)(n + 2) - n(n + 1)(n - 1)

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

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

Mk suýt nữa biết !

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

\(=\frac{1.2.3+2.3.3+3.4.3+...+n\left(n+1\right).3}{3}\)

\(=\frac{1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+n\left(n+1\right)\left[\left(n+2\right)-\left(n-1\right)\right]}{3}\)

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

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

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

3A = 1.2.3 + 2.3.3 + 3.4.3 + ..... + n(n + 1)3

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

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

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

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

3A=1.2.3+2.3.3+3.4.3+...+n(n+1).3

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

3A=1.2.3-1.2.0+2.3.4-2.3.3+.+n(n+1)+(n+2)-(n-1)+n(n-1)

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

3A=1.2.3+2.3.3+3.4.3+.....+N(N+1).3

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

3a=1.2.3-1.2.0+2.3.4-2.3.3+....+n(n+1)+(n+2)-(n-1)+n(n+1)

=>n(n-1)+(n+2)=n(n-1)+(n+2)/3

3A=1.2.3+2.3.3+3.4.3+......+N(N+1).3

3A=1.2.(3-0)+2.3.(4-1)+3.4.(5-2)+....+N(N+1)(N-2)-(N-1)

3A=1.2.3-1.2.0+2.3.4-2.3.1+......+N(N+1)+(N+2)-N-1+N(N-1)

=>N(N-1)+(N+2)=N(N-1)-(N+2)/3

cau hỏi tương tự ko có mà!!!!!!!!!!!!!!!!!!!!!!!!!!!!

3C=1.2.3+2.3.(4-1)+3.4.(5-2)+...+2014.2015.(2016-2013)

3C=2014.2015.2016

C=2014.2015.2016:3

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

VT=VP=>đpcm

b)áp dụng a)

\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+..+\frac{1}{99}-\frac{1}{100}=\frac{1}{1}-\frac{1}{100}=\frac{100}{100}-\frac{1}{100}=\frac{99}{100}\)

Vậy A=99/100

b) A=1-1/2+1/2-1/3+1/3-1/4+...+1/99-1/100

=1-1/100

=99/100

=9,9

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