Question

[nguyễn ngọc linh]

 a, S = \(\dfrac{1}{2}\) + \(\dfrac{1}{2^2}\) + \(\dfrac{1}{2^3}\) + ... + \(\dfrac{1}{2^{50}}\)

 b, S = \(\dfrac{1}{3}\) + \(\dfrac{1}{3^2}\) + \(\dfrac{1}{3^3}\) + ... + \(\dfrac{1}{3^{20}}\)

 c, S = \(\dfrac{1}{2}\) + \(\dfrac{1}{2^2}\) + \(\dfrac{1}{2^3}\) + ... + \(\dfrac{1}{2^{2024}}\)

 d, S = \(\dfrac{1}{3}\) + \(\dfrac{1}{3^2}\) + \(\dfrac{1}{3^3}\) + ... + \(\dfrac{1}{3^{100}}\)

Answer 1

[Akai Haruma]

a.

$S=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{50}}$

$2S=1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{49}}$

$\Rightarrow 2S-S=1-\frac{1}{2^{50}}$

$\Rightarrow S=1-\frac{1}{2^{50}}$

b.

$S=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{20}}$

$3S=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{19}}$

$\Rightarrow 3S-S=1-\frac{1}{3^{20}}$

$\Rightarrow 2A=1-\frac{1}{3^{20}}$

$\Rightarrow A=\frac{1}{2}-\frac{1}{2.3^{20}}$

Answer 2

[Lưu Nguyễn Hà An]

@Trần Bảo Việt, bạn không trả lời linh tinh nhé!