Question

Tính B= 1.2.3+2.3.4+...+(n-1) n (n+1) 

 HELP ME

 AI NHANH NHẤT MK TK CHO !

Answer 1

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

Answer 2

xin loi ban toaan lop 6 ban a

Answer 3

B=1.2.3+2.3.4+...+(n-1)n(n+1) 

4B=4.1.2.3+4.2.3.4+...+4(n-1)n(n+1) 

4B=(4-0)1.2.3+(5-1)2.3.4+...+[(n+2)-(n-2)](n-1)n(n+1) 

4B=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)

4B=(n-1)n(n+1)(n+2)

B=\(\frac{\left(n-1\right)n\left(n+1\right)\left(n+2\right)}{4}\)