A  = 1.2.3 + 2.3.4 + ....+ 48.49.50

=> 4A = 1.2.3.4 + 2.3.4.(5-1) + ...+ 48.49.50.(51-17)

= 1.2.3.4 + 2.3.4.5 - 1.2.3.4 + .....+ 48.49.50.51 - 47.48.49.50

= 48.49.50.51

=> A =  48.49.50.51:4 = 12.49.50.51

bài b) làm tương tự nha

- Với \(n=1\Rightarrow1.2.3=\frac{1.2.3.4}{4}\) (đúng)

- Giả sử biểu thức đúng với \(n=k\) hay:

\(1.2.3+...+k\left(k+1\right)\left(k+2\right)=\frac{k\left(k+1\right)\left(k+2\right)\left(k+3\right)}{4}\)

Ta cần chứng minh nó đúng với \(n=k+1\) hay:

\(1.2.3+...+k\left(k+1\right)\left(k+2\right)+\left(k+1\right)\left(k+2\right)\left(k+3\right)=\frac{\left(k+1\right)\left(k+2\right)\left(k+3\right)\left(k+4\right)}{4}\)

Thật vậy, ta có:

\(1.2.3+...+k\left(k+1\right)\left(k+2\right)+\left(k+1\right)\left(k+2\right)\left(k+3\right)\)

\(=\frac{k\left(k+1\right)\left(k+2\right)\left(k+3\right)}{4}+\left(k+1\right)\left(k+2\right)\left(k+3\right)\)

\(=\left(k+1\right)\left(k+2\right)\left(k+3\right)\left[\frac{k}{4}+1\right]\)

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

\(=\frac{\left(k+1\right)\left(k+2\right)\left(k+3\right)\left(k+4\right)}{4}\) (đpcm)

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

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

\(2A=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)

\(2A=\frac{1}{1.2}-\frac{1}{\left(n+1\right)\left(n+2\right)}\Rightarrow A=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right)\)

mình áp dụng công thức tổng quát:\(\frac{a}{n\left(n+1\right)\left(n+2\right)...\left(n+a\right)}=\frac{1}{n\left(n+1\right)\left(n+a-1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)...\left(n+a\right)}\)

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

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

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

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

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

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

tổng quát:  1/n(n+1)(n+2)=1/2[1/n(n+1) - 1/(n+1)(n+2)]

ai k dung mik giai cho

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

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

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

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

P/S:  tham khảo nhé

đến đây bn làm tiếp nha

Lời giải: Sử dụng hằng đẳng thức \(\frac{2}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)  ta có:

S_n=\(\frac{1}{2}\left[\frac{1}{1\times2}-\frac{1}{2\times3}\right]+\frac{1}{2}\left[\frac{1}{2\times3}-\frac{1}{3\times4}\right]+...\)\(+\frac{1}{2}\left[\frac{1}{\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right]\)

\(=\frac{1}{2}\left[\frac{1}{1\times2}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right]=\frac{n\left(n+3\right)}{4\left(n+1\right)\left(n+2\right)}\)

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

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

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

\(B=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+......+\dfrac{1}{n\left(n+1\right)\left(n+2\right)}\)

\(=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+....+\dfrac{1}{n\left(n+1\right)}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\)

\(=\dfrac{1}{1.2}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\)

\(=\dfrac{1}{2}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\)

Vậy..

\(B=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{n\left(n+1\right)}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\right)\)

\(B=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\right)=\dfrac{n^2+3n+2-2}{4\left(n+1\right)\left(n+2\right)}=\dfrac{n\left(n+3\right)}{4\left(n+1\right)\left(n+2\right)}\)

2B=\(\dfrac{2}{1.2.3}\)+\(\dfrac{2}{2.3.4}\)+....+\(\dfrac{2}{n\left(n+1\right)\left(n+2\right)}\)

=\(\dfrac{1}{1.2}\)-\(\dfrac{1}{2.3}\)+\(\dfrac{1}{2.3}\)-\(\dfrac{1}{3.4}\)+...+\(\dfrac{1}{n\left(n+1\right)}\)-\(\dfrac{1}{\left(n+1\right)\left(n+2\right)}\)

=\(\dfrac{1}{2}\)-\(\dfrac{1}{n^2+3n+2}\)

=\(\dfrac{n^2+3n}{2\left(n^2+3n+2\right)}\)

=>B=\(\dfrac{n^2+3n}{2\left(n^2+3n+2\right)}\):2

đến đây bạn tự tính nha !

Đặt A là tên biểu thức

A=1.2.3+2.3.4+...+n(n+1)(n+2)

4A=1.2.3.4+2.3.4.4+...+n(n+1)(n+2).4

4A=1.2.3.4 - 0.1.2.3 + 2.3.4.5 - 1.2.3.4 +...+ n(n+1)(n+2)(n+3) - (n-1)n(n+1)(n+2)

4A=[1.2.3.4+2.3.4.5+...+n(n+1)(n+2)(n+3)] - [0.1.2.3+1.2.3.4+...+(n-1)n(n+1)(n+2)]

4A=n(n+1)(n+2)(n+3)-0.1.2.3

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

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

\(\Rightarrow4A=1.2.3.4+2.3.4.4+3.4.5.4+...+4n\left(n+1\right)\left(n+2\right)\)

\(\Rightarrow4A=1.2.3.4+1.2.3.\left(5-1\right)+...+n\left(n+1\right)\left(n+2\right)\left(n+3-n+1\right)\)

\(\Rightarrow4A=1.2.3.4+2.3.4.5-1.2.3.4+...+n\left(n+1\right)\left(n+2\right)\left(n+3\right)-\left(n-1\right)n\left(n+1\right)\left(n\right)\)

\(\Rightarrow4A=n\left(n+1\right)\left(n+2\right)\left(n+3\right)\)

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