Question

Bài 1:

Cho A=\(\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{50^2}\)

Chứng minh A<2

Bài 2:

Tính tổng: S=\(3+\dfrac{3}{2}+\dfrac{3}{2^2}+...+\dfrac{3}{2^9}\)

Giúp mình nha

Answer 1

1)Ta thấy: \(\dfrac{1}{n^2}=\dfrac{1}{n.n}< \dfrac{1}{\left(n-1\right)n}\)

=>A=\(\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}...+\dfrac{1}{50^2}< 1+\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{49.50}\)

A<\(1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}=2-\dfrac{1}{50}< 2\)

Vậy A<2

2)Ta có:2S=6+3+\(\dfrac{3}{2}+\dfrac{3}{2^2}+...+\dfrac{3}{2^8}\)

2S-S=(6+3+\(\dfrac{3}{2}+\dfrac{3}{2^2}+...+\dfrac{3}{2^8}\))-(3+\(\dfrac{3}{2}+\dfrac{3}{2^2}+...+\dfrac{3}{2^9}\))

=>S=6-\(\dfrac{3}{2^9}=\dfrac{6.2^9-3}{2^9}\)

Vậy S=\(\dfrac{6.2^9-3}{2^9}\)

Answer 2

Các bạn cố giúp mink nhé mai mình phải nộp rồi

Answer 3

Lời giải:

Câu 1:

Dễ thấy:

\(\dfrac{1}{1^2}=1\)

\(\dfrac{1}{2^2}=\dfrac{1}{2.2}< \dfrac{1}{1.2}\)

\(\dfrac{1}{3^2}=\dfrac{1}{3.3}< \dfrac{1}{2.3}\)

\(....\)

\(\dfrac{1}{50^2}=\dfrac{1}{50.50}< \dfrac{1}{49.50}\)

\(\Rightarrow A< 1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\)

\(\Rightarrow A< 1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\)

\(\Rightarrow A< 1+1-\dfrac{1}{50}\)

\(\Rightarrow A< 2-\dfrac{1}{50}\)

\(\Rightarrow A< 2\)

Câu 2:

Ta có: \(S=3+\dfrac{3}{2}+\dfrac{3}{2^2}+\dfrac{3}{2^3}+...+\dfrac{3}{2^9}\)

\(\Rightarrow2S=6+3+\dfrac{3}{2}+\dfrac{3}{2^2}+\dfrac{3}{2^2}+...+\dfrac{3}{2^8}\)

\(\Rightarrow2S-S=S=6-\dfrac{3}{2^9}=6-\dfrac{3}{512}=\dfrac{3069}{512}\)

\(\Rightarrow S=\dfrac{3069}{512}\)