Tính tổng
B=1.2.3+2.3.4+3.4.5+...+n (n+1).(n+2)
Với n € N*
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) = \(\frac{\left(n-1\right).n.\left(n+1\right).\left(n+2\right)}{4}\)
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
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
tính tổng 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)
={1.2.3.(4-0)+2.3.4(5-1)+3.4.5.(6-2)+...+n(n+1)(n+2)[(n+3)-(n-1)]} : 4
= [1.2.3.4+2.3.4.5+3.4.5.6+...+n(n+1)(n+2)(n+3) - 1.2.3.4 - 2.3.4.5 - 3.4.5.6 - ... - n(n+1)(n+2)(n-1)] : 4
=\(\frac{\text{ n(n+1)(n+2)(n+3) }}{4}\)
B = \(\frac{n\left(n+1\right)\left(n+2\right)\left(n+3\right)}{4}\)
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 !!! :/
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
tính S = 1.2.3+2.3.4+3.4.5+...+n.(n+1).(n+2)
Ta có: S = 1.2.3+2.3.4+3.4.5+...+n.(n+1).(n+2)
=> 4 S = 1.2.3.(4-0) + 2.3.4.( 5-1) +........+ n.(n+1). (n+2). ((n+3)- (n-1))
= 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)
= n.(n+1). (n+2).(n+3)
Tính
N = 1/1.2.3 + 1/2.3.4 + 1/3.4.5 +...+ 1/n.(n+1).(n+2)
\(N=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{n.\left(n+1\right).\left(n+2\right)}\)
\(\Rightarrow2N=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{n.\left(n+1\right).\left(n+2\right)}\)
\(\Rightarrow N=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{n.\left(n+1\right)}-\frac{1}{\left(n+1\right).\left(n+2\right)}\right)\)
\(\Rightarrow N=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{\left(n+1\right).\left(n+2\right)}\right)\)
N=1/1.2.3 +1/2.3.4 +1/3.4.5 +...+1/n.(n+1).(n+2)
⇒2N=2/1.2.3 +2/2.3.4 +2/3.4.5 +...+2/n.(n+1).(n+2)
⇒N=1/2 .(1/1.2 −1/2.3 +1/2.3 −1/3.4 +1/3.4 −1/4.5 +...+1/n.(n+1) −1/(n+1).(n+2) )
⇒N=1/2 .(1/1.2 −1/(n+1).(n+2) )
chúc bạn học tốt !