Question

Tính:

a,   1.2.3+2.3.4+3.4.5+.................+n.(n+1).(n+2).

b,   1.3+2.4+3.5+4.6+..........................+(n-1).(n+1).

c,   1²+2²+3²+4²+......................+2014²+2015².

Answer 1

A=1.2.3+2.3.4+3.4.5+...+n(n+1)(n+2)

suy ra 4A=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)

Answer 2

Đặt ak  = k.(k+1).(k+2)

4a1 = 1.2.3.3-0.1.2.3

4a2 = 2.3.4.3-1.2.3.3

     ………….

4an-1 = (n-1).n.(n+1).(n+2)-(n-2).(n-1).n.(n+1)

4an = n.(n+1).(n+2).(n+3)-(n-1).n.(n+1).(n+2)

     Cộng từng vế n, ta được:

4(a1+a2+a3+………….+an) = n.(n+1).(n+2).(n+3)

4[1.2.3+2.3.4+3.4.5+………………..+n.(n+1).(n+2)] = n.(n+1).(n+2).(n+3)

=> A =   \(\frac{n.\left(n+1\right).\left(n+2\right).\left(n+3\right)}{4}\)