Tinh nhanh
A= \(1.2.3+2.3.4+3.4.5+...+48.49.50\)
B = \(1.2.3+2.3.4+3.4.5+...+n.\left(n+1\right).\left(n+2\right)\)
P = 1/1.2.3 + 1/2.3.4 + 1/3.4.5 +...+ 1/n(n+1)(n+2)
S = 1/1.2.3 + 1/2.3.4 + 1/3.4.5 +...+ 1/48.49.50 .
tao có:
2p=2/1.2.3+2/2.3.4+...+2/n.n(+1)n(n+2)
2p=3-1/1.2.3+4-2/1.2.3+...+(n+2)-n/n.(n+1).(n+2)
2p=3/1.2.3-1/1.2.3+4/2.3.4-2/2.3.4+...+(n+2)/n.(n+1).(n+2)-n/n.(n+1).(n+2)
2p=1/1.2-1/2.3+1/2.3-1/3.4+...+1/n.(n+1)-1/(n+1).(n+2)
2p=1/1.2-1/(n+1).(n+2)
2p=(n+!).(n+2)-2/(2n+2).(n+2)
suy ra p=(n+1).(n+2)-2/(2n+2).(2n+4)
2s=3-1/1.2.3+4-2/1.2.3+...+50-48/48.49.50
2s=3/1.2.3-1/1.2.3+4/2.3.4-2/2.3.4+...+50/49.50.48-48/48.50.49
2s=1/1.2-1/2.3+1/2.3-1/3.4+...+1/48.49-1/49.50
2s=1/1.2-1/49.50
'2s=1/2-1/2450
2s=1225/2450-1/2450
2s=1224/2450
s=612/1225
\(P=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)}\)1
\(P=\frac{1}{2}\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+...+\frac{2}{n\left(n+1\right)\left(n+2\right)}\right)\)
\(P=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+...+\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right)\)
\(P=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right)\)
\(P=\frac{\left(\frac{1}{2}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right)}{2}\)
S cx tinh giong v
1.2.3+2.3.4+3.4.5+...+n(n+1)(n+2)=\(\frac{n\left(n+1\right)\left(n+2\right)\left(n+3\right)}{4}\)
- 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)
Tính giá trị của tổng sau : \(1.2.3+2.3.4+3.4.5+...+n\left(n+1\right)\left(n+2\right)\)
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=\(\frac{n\left(n+1\right)\left(n+2\right)\left(n+3\right)}{4}\)
\(A=1.2.3+2.3.4+3.4.5+...+n\left(n+1\right)\left(n+2\right)\)
\(\Rightarrow4A=1.2.3.4+2.3.4.4+3.4.5.4+...+4n\left(n+1\right)\left(n+2\right)\)
\(\Rightarrow4A=1.2.3.4+1.2.3.\left(5-1\right)+...+n\left(n+1\right)\left(n+2\right)\left(n+3-n+1\right)\)
\(\Rightarrow4A=1.2.3.4+2.3.4.5-1.2.3.4+...+n\left(n+1\right)\left(n+2\right)\left(n+3\right)-\left(n-1\right)n\left(n+1\right)\left(n\right)\)
\(\Rightarrow4A=n\left(n+1\right)\left(n+2\right)\left(n+3\right)\)
\(\Rightarrow A=\frac{n\left(n+1\right)\left(n+2\right)\left(n+3\right)}{4}\)
Tính tổng \(B=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{n\left(n+1\right)\left(n+2\right)}\)
\(B=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+......+\dfrac{1}{n\left(n+1\right)\left(n+2\right)}\)
\(=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+....+\dfrac{1}{n\left(n+1\right)}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\)
\(=\dfrac{1}{1.2}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\)
\(=\dfrac{1}{2}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\)
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 !
Tính:
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)}\)
\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)}\)
\(2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{n\left(n+1\right)\left(n+2\right)}\)
\(2A=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)
\(2A=\frac{1}{1.2}-\frac{1}{\left(n+1\right)\left(n+2\right)}\Rightarrow A=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right)\)
mình áp dụng công thức tổng quát:\(\frac{a}{n\left(n+1\right)\left(n+2\right)...\left(n+a\right)}=\frac{1}{n\left(n+1\right)\left(n+a-1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)...\left(n+a\right)}\)
Tính:
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)}\))
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)]
Tính tổng
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{n.\left(n+1\right).\left(n+2\right)}\)
Đặt C =\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)}\)
\(\Rightarrow2C=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{n\left(n+1\right)\left(n+2\right)}\)
\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)
\(=\frac{1}{1.2}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)
\(\Rightarrow C=\left(\frac{1}{1.2}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right)\div2\)
Tính
D=\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+......+\frac{1}{n\left(n+1\right)\left(n+2\right)}\)
\(D=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)}\)
\(=\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{n\left(n+1\right)\left(n+2\right)}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right)\)
P/S: tham khảo nhé
đến đây bn làm tiếp nha
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{n.\left(n+1\right).\left(n+2\right)}\)
chúng ta hãy quy đồng rồi cộng chúng lại với nhau thì sẽ ra kết quả và cậu hãy xem lai kiến thức mới học của cậu đi