Cho dãy số \(\left( {{u_n}} \right)\) được xác định như sau: \(\left\{ \begin{array}{l}{u_1} = 2,{u_2} = 3\\{u_{n + 2}} = 2{u_{n + 1}} - {u_n}\end{array} \right.\left( {\forall n \in {\mathbb{N}^*},n \ge 2} \right)\). Khi đó tổng \({u_1} + {u_2} + ... + {u_n}\) bằng:

Ta có:

\(\begin{array}{l}{u_3} = 2{u_2} - {u_1}\\{u_4} = 2{u_3} - {u_2}\\{u_5} = 2{u_4} - {u_3}\\...\\{u_{n + 1}} = 2{u_n} - {u_{n - 1}}\end{array}\)

Cộng vế với vế ta được:

\(\begin{array}{l}{u_3} + {u_4} + ... + {u_{n + 1}} = 2{u_n} + {u_{n - 1}} + ... + {u_3} + {u_2} - {u_1}\\ \Leftrightarrow {u_n} + {u_{n + 1}} = 2{u_n} + {u_2} - {u_1}\\ \Leftrightarrow {u_{n + 1}} = {u_n} + {u_2} - {u_1} = {u_n} + 3 - 2 = {u_n} + 1\end{array}\)

Ta lại có:

\(\begin{array}{l}{u_2} = {u_1} + 1\\{u_3} = {u_2} + 1 = \left( {{u_1} + 1} \right) + 1 = {u_1} + 2.1\\{u_4} = {u_3} + 1 = \left( {{u_1} + 2.1} \right) + 1 = {u_1} + 3.1\\...\\{u_n} = {u_{n - 1}} + 1 = \left( {{u_1} + \left( {n - 2} \right).1} \right) + 1 = {u_1} + \left( {n - 1} \right).1 = 2 + n - 1 = n + 1\end{array}\)

Vậy ta có:

\(\begin{array}{l}{u_1} + {u_2} + ... + {u_n} = 2 + 3 + ... + \left( {n + 1} \right) = \left[ {1 + 2 + 3 + ... + \left( {n + 1} \right)} \right] - 1\\ = \frac{{\left( {n + 1} \right)\left( {n + 2} \right)}}{2} - 1 = \frac{{{n^2} + n + 2n + 2 - 2}}{2} = \frac{{{n^2} + 3n}}{2} = \frac{{n\left( {n + 3} \right)}}{2}\end{array}\)