B=1.2.3+2.3.4+3.4.5+..........+(n-1).n (n+1)
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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.
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
549 + X = 1326
X = 1326 - 549
X = 777
X - 636 = 5618
X = 5618 + 636
X = 6254
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) = \(\frac{\left(n-1\right).n.\left(n+1\right).\left(n+2\right)}{4}\)
4(1.2.3) = 1.2.3.4 - 0.1.2.3
4(2.3.4) = 2.3.4.5 - 1.2.3.4
4(3.4.5) = 3.4.5.6 - 2.3.4.5
....................................
4(n-1)n(n+1) = (n-1)n(n+1)(n+2) - (n-2)(n-1)n(n+1)
=> 4 B = (n-1)n(n+1)(n+2) => B= (n-1)n(n+1)(n+2):4
4(1.2.3)=1.2.3.4 - 0.1.2.3
4(2.3.4)=2.3.4.5 - 1.2.3.4
4(3.4.5)=3.4.5.6 - 2.3.4.5
.........................
......................................
.......................................
=\(\frac{\left(n-1\right)n\left(n+1\right)\left(n+2\right)}{4}\)
4B = 1.2.3.4 + 2.3.4.4 + ... + (n-1)n(n+1).4
= 1.2.3.4 - 0.1.2.3 + 2.3.4.5 - 1.2.3.4 + ... + (n-1)n(n+1)(n+2) - [(n-2)(n-1)n(n+1)]
= (n-1)n(n+1)(n+2) - 0.1.2.3
= (n-1)n(n+1)(n+2)
suy ra \(B = {(n-1)n(n+1)(n+2)\over 4}\)
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)
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
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 = 1.2.3 + 2.3.4 +..+ (n-1).n.(n+1)
4B = 1.2.3.4 + 2.3.4.4 + 3.4.5.4 +..+ (n-1)n(n+1).4
ghi dọc cho dễ nhìn:
(k-1)k(k+1).4 = (k-1)k(k+1)[(k+2) - (k-2)] = (k-1)k(k+1)(k+2) - (k-2)(k-1)k(k+1)
ad cho k chạy từ 2 đến n ta có:
1.2.3.4 = 1.2.3.4
2.3.4.4 = 2.3.4.5 - 1.2.3.4
3.4.5.4 = 3.4.5.6 - 2.3.4.5
...
(n-2)(n-1)n.4 = (n-2)(n-1)n(n+1) - (n-3)(n-2)(n-1)n
(n-1)n(n+1).4 = (n-1)n(n+1)(n+2) - (n-2)(n-1)n(n+1)
+ + cộng lại vế theo vế + + (chú ý cơ chế rút gọn)
4B = (n-1)n(n+1)(n+2)
=> B = [(n-1)n(n+1)(n+2)]/4