1.2.3+2.3.4+3.4.5+.........+n.(n+1).(n+2) = ?
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
Bài 4:
a) Chứng minh các công thức sau:
A = 1.2.3+2.3.4+3.4.5+...+(n-2)(n-1)n = (n−2).(n−1).n.(n+1):
4
b) Áp dụng tính tổng sau: G = 1.2.3 + 2.3.4 + 3.4.5 +...+ 2021.2022.2023
4A = 4.[1.2.3 + 2.3.4 + 3.4.5 + … + (n – 1).n.(n + 1)]
4A = 1.2.3.4 + 2.3.4.4 + 3.4.5.4 + … + (n – 1).n.(n + 1).4
4A = 1.2.3.4 + 2.3.4.(5 – 1) + 3.4.5.(6 – 2) + … + (n – 1).n.(n + 1).[(n + 2) – (n – 2)]
4A = 1.2.3.4 + 2.3.4.5 – 1.2.3.4 + 3.4.5.6 – 2.3.4.5 + … + (n – 1).n(n + 1).(n + 2) – (n – 2).(n – 1).n.(n + 1)
4A = (n – 1).n(n + 1).(n + 2)
A = (n – 1).n(n + 1).(n + 2) : 4.
cau a thi sao ha ban ?
ok thanks ban nhe
Câu5: Tính : 1.2.3+2.3.4+3.4.5+...................+28.29.30.Từ đó cho biết kết quả của tổng : 1.2.3+2.3.4+3.4.5+............................+(n-1).n.(n+1) theo n
(với n là số tự nhiên khác 0 )
Đặt A = 1.2.3 + 2.3.4 + 3.4.5 + ... + 28.29.30
4A = 1.2.3.(4-0) + 2.3.4.(5-1) + 3.4.5.(6-2) + ... + 28.29.30.(31-27)
4A = 1.2.3.4 - 0.1.2.3. + 2.3.4.5 - 1.2.3.4 + 3.4.5.6 - 2.3.4.5 + ... + 28.29.30.31 - 27.28.29.30
4A = 28.29.30.31 - 0.1.2.3
4A = 28.29.30.31
\(A=\frac{28.29.30.31}{4}=7.29.30.31=188790\)
Theo cách tính trên ta dễ dàng tính được:
1.2.3 + 2.3.4 + 3.4.5 + ... + (n - 1).n.(n + 1) = \(\frac{\left(n-1\right).n.\left(n+1\right).\left(n+2\right)}{4}\)
Tinh nhanh
A= \(1.2.3+2.3.4+3.4.5+...+48.49.50\)
B = \(1.2.3+2.3.4+3.4.5+...+n.\left(n+1\right).\left(n+2\right)\)
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
B=1.2.3+2.3.4+3.4.5+...+n(n+1)(n+2)
B=1.2.3+2.3.4+3.4.5+...+n.(n+1).(n+2)
4B=4.(1.2.3+2.3.4+3.4.5+...+n.(n+1).(n+2)
4B=1.2.3.4+2.3.4.4+3.4.5.4+...+n.(n+1).(n+2).4
4B=1.2.3.4+2.3.4.(5-1)+3.4.5.(6-2)+...+n.(n+1)+(n+2).(n+3-4)
4B=1.2.3.4+2.3.4.5-1.2.3.4+3.4.5.6-2.3.4.5+...+n.(n+1).(n+2).(n+3)-(n-1).n.(n+1).(n+2)
4B=(1.2.3.4-1.2.3.4)+(2.3.4.5-2.3.4.5)+...+[(n-1).n.(n+1).(n+2)-(n-1).n.(n+1).(n+2)]+n.(n+2).(n+2).(n+3)
4B=0+0+...+0+n.(n+1).(n+2).(n+3)
4B=n.(n+1).(n+2).(n+3)
B=[n.(n+1).(n+2).(n+3)]:4
Tính:
f) F= 1.2+2.3+3.4+...+n(n+1)
g) G= 1.2.3+2.3.4+3.4.5+...+99.100.101
h) H= 1.2.3+2.3.4+3.4.5+...+n(n+1)(n+2)
i) I= 1.3+2.4+3.5+...+99.100
j) J= 1.4+2.5+3.6+...+99.102
Tính : a, 1.2.3 + 2.3.4 + 3.4.5 + ... + (n - 1).n.(n+1)
b, 1.2.3 + 3.4.5 + 5.6.7 + 98.99.100
549 + X = 1326
X = 1326 - 549
X = 777
X - 636 = 5618
X = 5618 + 636
X = 6254
549 ,1326 ở đâu zậy bạn !!! :/
S=1.2.3+2.3.4+3.4.5+...+n.(n+1) 6+2 (n thuộc N*)
Tính tổng : 1.2 + 2.3 + 3.4 + …..+ n.(n+1)
1.2.3+ 2.3.4 + 3.4.5 + ….+ n(n+1)(n+2)
https://olm.vn/hoi-dap/tim-kiem?q=t%C3%ADnh+t%E1%BB%95ng+sau+:S+=+1.2.3+2.3.4+3.4.5+...+n.(n+1).(n+2)+&id=601088