Question

[trần tú trân]

Kí hiệu (x) là số nguyên lớn nhất không quá x

Tìm (a) biết 

a=\(\left(\frac{1}{2}\right)^2\)+\(\left(\frac{1}{3}\right)^2\)+...+\(\left(\frac{1}{2014}\right)^2\)

Answer 1

[Nhọ Nồi]

Kí hiệu sai, phải là [a]

+) Vì \(\left(\frac{1}{2}\right)^2>0;\left(\frac{1}{3}\right)^2>0;\left(\frac{1}{4}\right)^2>0;...;\left(\frac{1}{2014}\right)^2>0\)

\(\Rightarrow\left(\frac{1}{2}\right)^2+\left(\frac{1}{3}\right)^2+...+\left(\frac{1}{2014}\right)^2>0\)

\(\Rightarrow a>0^{\left(1\right)}\)

+) Ta có: \(\left(\frac{1}{2}\right)^2<\frac{1}{1.2};\left(\frac{1}{3}\right)^2<\frac{1}{2.3};...;\left(\frac{1}{2014}\right)^2<\frac{1}{2013.2014}\)

\(\Rightarrow\left(\frac{1}{2}\right)^2+\left(\frac{1}{3}\right)^2+...+\left(\frac{1}{2014}\right)^2<\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2013.2014}\)

\(\Rightarrow a<1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2013}-\frac{1}{2014}\)

\(\Rightarrow a<1-\frac{1}{2014}<1^{\left(2\right)}\)

Từ \(^{\left(1\right)}\) và \(^{\left(2\right)}\) => 0 < a < 1

=> [a] = 0