Question

Tìm số nguyên n, biết rằng:

\(\dfrac{1}{4} . \dfrac{2}{6} . \dfrac{3}{8} .\dfrac{4}{10} . \dfrac{5}{12} .... \dfrac{30}{62} . \dfrac{31}{64} = 2^{n}\)\(\)

Answer 1

\(\dfrac{1}{2.2}.\dfrac{2}{2.3}.....\dfrac{31}{64}=2^x\\ =>\dfrac{1}{2.2.2.....2.64}=2^x\\ \dfrac{1}{2^{30}.26}=2^x\\ =>\dfrac{1}{2^{36}}=2^x\\ =>2^{-36}=2^x\\ =>x=-36\)

Answer 2

\(n=-36\)

Answer 3

Ta có:  \(2^n=\dfrac{1}{4}.\dfrac{2}{6}.\dfrac{3}{8}.\dfrac{4}{10}.\dfrac{5}{12}....\dfrac{30}{62}.\dfrac{31}{64}\)

⇔ \(2^n=\dfrac{1.2.3.4....31}{2.\left(2.3.4.....1\right).64}=\)

⇔ \(2^n=\dfrac{1}{2}.\dfrac{1}{64}=\dfrac{1}{128}\)  \(\Leftrightarrow\)   \(2^n=\dfrac{1}{2^6}\)

⇔ \(2^{x+6}=1\)

⇔ \(x+6=0\)

⇒  \(\left\{{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\)