Câu hỏi của Kirigaya Kazuto

Question

Cho biểu thức :

$A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2014}}$

Hãy so sánh A với $\dfrac{3}{2}$

Lightning Farron · Accepted Answer

$A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2014}}$

$3A=3\left(1+\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2014}}\right)$

$3A=3+1+\dfrac{1}{3}+...+\dfrac{1}{3^{2013}}$

$3A-A=\left(3+1+...+\dfrac{1}{3^{2013}}\right)-\left(1+\dfrac{1}{3}+...+\dfrac{1}{3^{2014}}\right)$

$2A=3-\dfrac{1}{3^{2014}}\Rightarrow A=\dfrac{3}{2}-\dfrac{\dfrac{1}{3^{2014}}}{2}< \dfrac{3}{2}$

Vậy $A< \dfrac{3}{2}$Câu trả lời: $A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2014}}$

$3A=3\left(1+\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2014}}\right)$

$3A=3+1+\dfrac{1}{3}+...+\dfrac{1}{3^{2013}}$

$3A-A=\left(3+1+...+\dfrac{1}{3^{2013}}\right)-\left(1+\dfrac{1}{3}+...+\dfrac{1}{3^{2014}}\right)$

$2A=3-\dfrac{1}{3^{2014}}\Rightarrow A=\dfrac{3}{2}-\dfrac{\dfrac{1}{3^{2014}}}{2}< \dfrac{3}{2}$

Vậy $A< \dfrac{3}{2}$

chugialinh · Answer

A=1+13+132+133+...+132014A=1+13+132+133+...+132014

3A=3(1+13+132+133+...+132014)3A=3(1+13+132+133+...+132014)

3A=3+1+13+...+1320133A=3+1+13+...+132013

3A−A=(3+1+...+132013)−(1+13+...+132014)3A−A=(3+1+...+132013)−(1+13+...+132014)

2A=3−132014⇒A=32−1320142<32Câu trả lời: A=1+13+132+133+...+132014A=1+13+132+133+...+132014

3A=3(1+13+132+133+...+132014)3A=3(1+13+132+133+...+132014)

3A=3+1+13+...+1320133A=3+1+13+...+132013

3A−A=(3+1+...+132013)−(1+13+...+132014)3A−A=(3+1+...+132013)−(1+13+...+132014)

2A=3−132014⇒A=32−1320142<32

Ôn tập toán 6