Cho dãy số (u_n) với u₁ = 2,$u_{n+1}=u_n+\frac{2}{3^n}$ , n ≥ 1. Đặt v_n = u_{n + 1} – u_n.

a) Tính v₁ + v₂ + ... + v_n theo n.

a) Ta có v_n = u_{n + 1} – u_n = $\left(u_n+\frac{2}{3^n}\right)-u_n=\frac{2}{3^n}$ .

Do đó, v₁ + v₂ + ... + v_n = $\frac{2}{3}+\frac{2}{3^2}+\mathrm{...}+\frac{2}{3^n}=2\left(\frac{1}{3}+\frac{1}{3^2}+\mathrm{...}+\frac{1}{3^n}\right)$

$=2.\frac{1-\frac{1}{3^{n+1}}}{1-\frac{1}{3}}=3\left(1-\frac{1}{3^{n+1}}\right)$.