Question

1, a, Tính (2 cách)

A=1x2+2x3+3x4+....+nx(n+1)

b, Nêu cách tính tổng quát

c, áp dụng tính

B=1x2x3+2x3x4+....+(n-1)x(n+1)

giúp mình với đang cần gấp

Answer 1

a)

\(A=1.2+2.3+3.4+...+n.\left(n+1\right)\)

\(3A=1.2.3+2.3.3+3.4.3+...+n.\left(n+1\right).3\)

\(3A=1.2.\left(3-0\right)+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+n.\left(n+1\right).\left[\left(n+2\right)-\left(n-1\right)\right]\)

\(3A=(1.2.3-0.1.2)+\left(2.3.4-1.2.3\right)+\left(3.4.5-2.3.5\right)+...+\left[n.\left(n+1\right).\left(n+2\right)-\left(n-1\right).n.\left(n+1\right)\right]\)\(3A=-0.1.2+n.\left(n+1\right).\left(n+2\right)\)

\(3A=n.\left(n+1\right).\left(n+2\right)\)

\(A=\dfrac{n.\left(n+1\right).\left(n+2\right)}{3}\)

Answer 2

c)

\(B=1.2.3+2.3.4+...+\left(n-1\right).n.\left(n+1\right)\)

\(4B=1.2.3.4+2.3.4.4+3.4.5.4+...+\left(n-1\right).n.\left(n+2\right).4\)

\(4B=1.2.3.4+2.3.4.\left(5-1\right)+3.4.5.\left(6-2\right)+...+\left(n-1\right).n.\left(n+1\right).\left[\left(n+2\right)-\left(n-2\right)\right]\)\(4B=1.2.3.4+\left(2.3.4.5-1.2.3.4\right)+\left(3.4.5.6-2.3.4.5\right)+...+\left[\left(n-1\right).n.\left(n+1\right).\left(n+2\right)-\left(n-1\right).n.\left(n+1\right).\left(n-2\right)\right]\)\(4B=\left(n-1\right).n.\left(n+1\right).\left(n+2\right)\\ B=\dfrac{\left(n-1\right).n.\left(n+1\right).\left(n+2\right)}{4}\)