Question

a) Tìm \(n\in N\)* :

+) \(\dfrac{1}{8}.16^n=2^n\)

+) \(27< 3^n< 243\)

b) Tính : \(\left(\dfrac{1}{4.9}+\dfrac{1}{9.14}+\dfrac{1}{14.19}+...+\dfrac{1}{44.49}\right).\dfrac{1-3-5-7-...-49}{89}\)

Answer 1

Bài 1 :

a) +) \(\dfrac{1}{8}\cdot16^n=2^n\)

\(\Leftrightarrow\dfrac{1}{8}=\dfrac{2^n}{16^n}\)

\(\Rightarrow\dfrac{1}{8}=\dfrac{1}{8}^n\)

Vậy n = 1.

+) \(27< 3^n< 243\)

\(\Leftrightarrow3^3< 3^n< 3^5\)

Vậy n = 4.

Bài 2 : \(\left(\dfrac{1}{4\cdot9}+\dfrac{1}{9\cdot14}+\dfrac{1}{14\cdot19}+...+\dfrac{1}{44\cdot49}\right)\cdot\dfrac{1-3-5-7-...-49}{89}\)

\(\Leftrightarrow\left(\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{19}+...+\dfrac{1}{44}-\dfrac{1}{49}\right)\cdot\dfrac{-623}{89}\)

\(\Leftrightarrow\left(\dfrac{1}{4}-\dfrac{1}{49}\right)\cdot\dfrac{-623}{89}=-\dfrac{45}{28}\)