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B = 1.2.3 + 2.3.4 + ... + (n - 1)n(n + 1)
=> 4B = 1.2.3.4 + 2.3.4.4 + (n-1).n(n+1).4
=> 4B =1.2.3.4 + 2.3.4.(5-1) + (n-1)n(n+1)(n+2-n+2)
=> 4B = 1.2.3.4+ 2.3.4.5- 1.2.3.4 + .... - (n-2)(n-1)n(n+1) + (n-1)n(n+1)(n+2)
=> 4B = (n-1)n(n+1)(n+2)
=> B = \(\frac{\left(n-1\right)n\left(n+1\right)\left(n+2\right)}{4}\)
Ta có: B = 1.2 + 2.3 + 3.4 + … + n.(n + 1)
=> 3A = 1.2.(3-0) + 2.3.(4-1) + .... + n.(n+1).(n+2 - n+1)
=> 3A = 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + .... + n.(n+1).(n+2)
=> 3A = n.(n+1).(n+2)
= > A =
S=1.2+2.3+3.4+.............+n(n+1)
=1(1+1) + 2(2+1) + 3(3+1) +...+n(n+1)
=(1^2 + 2^2 + 3^2 +...+ n^2) + (1 + 2 + 3 + ...+ n)
Ta có các công thức:
1^2 + 2^2 + 3^2 +...+ n^2 = n(n+1)(2n+1)/6
1 + 2 + 3 + ...+ n = n(n+1)/2
Thay vào ta có:
S = n(n+1)(2n+1)/6 + n(n+1)/2
=n(n+1)/2[(2n+1)/3 + 1]
=n(n+1)(n+2)/3
B = 1.2.3 + 2.3.4 + ... + (n - 1)n(n + 1)
4B = 1.2.3.4 + 2.3.4.4 + ... + (n - 1)n(n + 1).4
4B = 1.2.3.4 + 2.3.4.(5 - 1) + 3.4.5.(6 - 2) + .... + (n - 1).n.(n + 1).[(n + 2) - (n - 2)]
4B = 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)
4B = (n-1)n(n+1)(n+2)
B = (n-1)n(n+1)(n+2) : 4
Ta có : 4B =4 . ( 1.2.3 + 2.3.4 + ...+ (n - 1 )n( n + 1 )
<=> 4B = 1.2.3 .( 4 - 0 ) + 2.3.4 .( 5- 1 ) + ... + ( n - 1 ) n ( n + 1 ) [ ( n + 2 ) - ( n - 2 ) ]
<=> 4B = 1 . 2 . 3 . 4 +2 . 3. 4 .5 -1.2.3 .4 + ... + ( n- 1 ) n ( n + 1 ) ( n + 2 )- ( n-1)( n+1).n/( n- 2 )
<=> 4B = ( n- 1 ).( n+1 ).n.( n + 2 )
<=> B = \(\frac{\left(n-1\right)\left(n+1\right)n\left(n+2\right)}{4}\)
Vậy B = \(\frac{\left(n-1\right)\left(n+1\right)n\left(n+2\right)}{4}\)
\(B=1.2.3+2.3.4+...+\left(n-1\right).n.\left(n+1\right)\)
\(4B=1.2.3.4+2.3.4.\left(5-1\right)+...+\left(n-1\right).n.\left(n+1\right)\left[\left(n+2\right)-\left(n-2\right)\right]\)
\(4B=1.2.3.4+2.3.4.5-1.2.3.4+...+\left(n-1\right).n.\left(n+1\right)\left(n+2\right)-\left(n-2\right)\left(n-1\right).n.\left(n+1\right)\)
\(4B=\left(n-1\right).n.\left(n+1\right)\left(n+2\right)\)
\(B=\frac{\left(n-1\right).n.\left(n+1\right)\left(n+2\right)}{4}\)
Tham khảo nhé~
Ta có: \(B=1.2.3+2.3.4+...+\left(n-1\right).n.\left(n+1\right)\)
\(\Leftrightarrow4B=4.\left[1.2.3+2.3.4+...+\left(n-1\right).n.\left(n+1\right)\right]\)
\(\Leftrightarrow4B=1.2.3.4+2.3.4.4+...+\left(n-1\right).n.\left(n+1\right).4\)
\(\Leftrightarrow4B=1.2.3.4+2.3.4\left(5-1\right)+...+\left(n-1\right)n.\left(n+1\right).\left[\left(n+2\right)-\left(n-2\right)\right]\)
\(\Leftrightarrow4B=1.2.3.4+2.3.4.5-1.2.3.4+...+\left(n-1\right).n.\left(n+1\right).\left(n+2\right)-\left(n-2\right).\)\(\left(n-1\right).n.\left(n+1\right)\)
\(\Leftrightarrow4B=\left(n-1\right).n.\left(n+1\right).\left(n+2\right)\)
\(\Leftrightarrow B=\left(n-1\right).n.\left(n+1\right).\left(n+2\right)\div4\)
Vậy \(B=\left(n-1\right).n.\left(n+1\right).\left(n+2\right)\div4\)
B=1*2*3+2*3*4+3*4*5+...+(n-1)n(n+1)
4B=1*2*3*4+2*3*4*(5-1)+3*4*5*(6-2)+...+(n-1)*n*(n+1)*[(n+2)-(n-2)]
4B=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)
4B=(n-1)n(n+1)(n+2)
B=[(n-1)n(n+1)(n+2)]:4
Nho k cho minh voi nha
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) = \(\frac{\left(n-1\right).n.\left(n+1\right).\left(n+2\right)}{4}\)
3F= 1.2.(3-0)+ 2.3.(4-1)+...+ n.(n+1).[(n+2)-(n-1)]
=[1.2.3+ 2.3.4+...+ (n-1)n(n+1)+ n(n+1)(n+2)]- [0.1.2+ 1.2.3+...+(n-1)n(n+1)]
=n(n+1)(n+2)
=>F
H=1.2.3+2.3.4+3.4.5+...+n(n+1)(n+2)
=> 4H=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)
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)
=> \(B=\frac{\left(n-1\right)n\left(n+1\right)\left(n+2\right)}{4}\)
k cho mik nha!
