Question

[Nguyễn Hà My]

Biết biểu thức 

 \(P=\sqrt{\dfrac{1}{4}+\dfrac{1}{1^2}+\dfrac{1}{3^2}}+\sqrt{\dfrac{1}{4}+\dfrac{1}{3^2}+\dfrac{1}{5^2}}+...+\sqrt{\dfrac{1}{4}+\dfrac{1}{399^2}+\dfrac{1}{401^2}}=\dfrac{a}{b};\)

, với a và b là các số nguyên dương, a/ b là phân số tối giản. Khi đó giá trị của biểu thức

Q= a  −100b bằng

A. 400 .                          B. 401.                           C. 403.                   D. 402 .

Answer 1

[Nguyễn Việt Lâm]

\(\sqrt{\dfrac{1}{4}+\dfrac{1}{\left(2n-1\right)^2}+\dfrac{1}{\left(2n+1\right)^2}}=\sqrt{\dfrac{\left(2n-1\right)^2\left(2n+1\right)^2+4\left(2n-1\right)^2+4\left(2n+1\right)^2}{4\left(2n-1\right)^2\left(2n+1\right)^2}}\)

\(=\sqrt{\dfrac{\left(4n^2-1\right)^2+4\left(4n^2-4n+1\right)+4\left(4n^2+4n+1\right)}{4\left(2n-1\right)^2\left(2n+1\right)^2}}\)

\(=\sqrt{\dfrac{16n^4+24n^2+9}{4\left(2n-1\right)^2\left(2n+1\right)^2}}=\sqrt{\dfrac{\left(4n^2+3\right)^2}{4\left(2n-1\right)^2\left(2n+1\right)^2}}=\dfrac{4n^2+3}{2\left(2n-1\right)\left(2n+1\right)}\)

\(=\dfrac{\left(4n^2-1\right)+4}{2\left(2n-1\right)\left(2n+1\right)}=\dfrac{1}{2}+\dfrac{2}{\left(2n-1\right)\left(2n+1\right)}\)

\(=\dfrac{1}{2}+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\)

Do đó:

\(P=\left(\dfrac{1}{2}+\dfrac{1}{1}-\dfrac{1}{3}\right)+\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}\right)+...+\left(\dfrac{1}{2}-\dfrac{1}{399}-\dfrac{1}{401}\right)\)

\(=\dfrac{1}{2}.200+1-\dfrac{1}{401}=\dfrac{40500}{401}\)

\(\Rightarrow Q=400\)