Tính A = 1.2 + 2.3 + 3.4 + … + n.(n + 1)

Đáp án:

Giải thích các bước giải: $\begin{array}{l}A{\rm{ }} = {\rm{ }}1.2{\rm{ }} + {\rm{ }}2.3{\rm{ }} + {\rm{ }}3.4{\rm{ }} + {\rm{ }} \ldots {\rm{ }} + {\rm{ }}n.\left( {n{\rm{ }} + {\rm{ }}1} \right)\\3.A = 1.2.3 + 2.3.3 + 3.4.3 + ... + n\left( {n + 1} \right).3\\ = 1.2.\left( {3 - 0} \right) + 2.3.\left( {4 - 1} \right) + ... + n.\left( {n + 1} \right).\left[ {\left( {n + 2} \right) - \left( {n - 1} \right)} \right]\\ = \left[ {1.2.3 + 2.3.4 + ... + \left( {n - 1} \right).n\left( {n + 1} \right)} \right] - \left[ {0.1.2 + 1.2.3 + ... + \left( {n - 1} \right).n.\left( {n + 1} \right)} \right]\\ = n.\left( {n + 1} \right).\left( {n + 2} \right)\\ \Rightarrow A = \frac{{n.\left( {n + 1} \right).\left( {n + 2} \right)}}{3}\end{array}$

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

=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)

=>S=[n.(n+1).(n+2)] /3