Đặt $S_n = \dfrac13 + \dfrac{1}{3^2} + \dfrac{1}{3^3} + \dots + \dfrac{1}{3^n} = \dfrac{1}{2}\cdot\dfrac{3^n - 1}{3^n}\qquad (*)$
- Với $n = 1$ ta có:
$\dfrac{1}{3} = \dfrac{1}{3}\quad $ (đúng)
- Giả sử $(*)$ đúng khi $n = k \geq 1$, tức là:
$S_k = \dfrac13 + \dfrac{1}{3^2} + \dfrac{1}{3^3} + \dots + \dfrac{1}{3^k} = \dfrac{1}{2}\cdot\dfrac{3^k - 1}{3^k}$ (giả thiết quy nạp)
- Ta cần chứng minh $(*)$ đúng với $n = k + 1$ tức là cần chứng minh:
$S_{k+1} = \dfrac13 + \dfrac{1}{3^2} + \dfrac{1}{3^3} + \dots + \dfrac{1}{3^k} + \dfrac{1}{3^{k+1}}= \dfrac{1}{2}\cdot\dfrac{3^{k+1} - 1}{3^{k+1}}$
Thật vậy:
$S_{k+1} = S_k + \dfrac{1}{3^{k+1}} = \dfrac{1}{2}\cdot\dfrac{3^k - 1}{3^k} + \dfrac{1}{3^{k+1}}$
$\to S_{k+1} = \dfrac{(3^k-1).3 + 2}{2.3^{k+1}} = \dfrac{1}{2}\cdot\dfrac{3^{k+1} - 1}{3^{k+1}}$
Vậy $\dfrac13 + \dfrac{1}{3^2} + \dfrac{1}{3^3} + \dots + \dfrac{1}{3^n} = \dfrac{1}{2}\cdot\dfrac{3^n - 1}{3^n}\qquad \forall n \in \Bbb N^*$
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Chứng minh theo cách trung học:
$S = \dfrac13 + \dfrac{1}{3^2} + \dfrac{1}{3^3} + \dots + \dfrac{1}{3^n}$
$\to 3S = 1 + \dfrac13 + \dfrac{1}{3^2} + \dfrac{1}{3^3} + \dots + \dfrac{1}{3^{n-1}}$
$\to S - 3S = \left(\dfrac13 + \dfrac{1}{3^2} + \dfrac{1}{3^3} + \dots + \dfrac{1}{3^n}\right) - \left(1 + \dfrac13 + \dfrac{1}{3^2} + \dfrac{1}{3^3} + \dots + \dfrac{1}{3^{n-1}}\right)$
$\to -2S = \dfrac{1}{3^n} - 1$
$\to S = \dfrac{1}{2}\cdot\dfrac{3^n - 1}{3^n}$
