Question

cmr:F=1.2.3+2.3.4+3.4.5+...+k.(k+1).(k+2)

4F+1 là số chính phương

Answer 1

 

4F = 1.2.3(4-0) + 2.3.4(5-1) +3.4.5.(6-2) +......+ k.(k+1)(k+2)[(k+3) - (k-1)]

    = 1.2.3.4 - 0 + 2.3.4.5 - 1.2.3.4 + 3.4.5.6 - 2.3.4.5 + ......+ k(k+1)(k+2)(k+3)- (k-1)k(k+1)(k+2) = k(k+1)(k+2)(k+3)

F = \(\frac{k\left(k+1\right)\left(k+2\right)\left(k+3\right)}{4}\)

=> 4F + 1 =\(k\left(k+1\right)\left(k+2\right)\left(k+3\right)+1\)=[k(k+3)][(k+1)(k+2)] +1 =(k²+3k)(k²+3k+2) + 1

               = (k²+3k)² +2(k²+3k) +1 = (k²+3k+1)²

=> 4F + 1 là số chình phương