\(Ta\) \(có\) :
\(S=\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{9999}{10000}\)\(=\frac{2^2-1}{2^2}+\frac{3^2-1}{3^2}+\frac{4^2-1}{4^2}+...+\frac{100^2-1}{100^2}\)
\(=99-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\right)\)
\(Đặt\) \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)
Do A > 0 nên S < 99 (1)
Do A\(=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(A< 1-\frac{1}{100}\)
Suy ra \(S=99-A>99-\left(1-\frac{1}{100}\right)\)
\(\Rightarrow S>98+\frac{1}{100}\Rightarrow S>98\) (2)
Lập luận ra điều phải chứng minh
\(S=\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{9999}{10000}\)
\(\Rightarrow S=\left(1-\frac{1}{4}\right)+\left(1-\frac{1}{9}\right)+\left(1-\frac{1}{16}\right)+...+\left(1-\frac{1}{10000}\right)\)
\(\Rightarrow S=1-\frac{1}{4}+1-\frac{1}{9}+1-\frac{1}{16}+...+1-\frac{1}{10000}\)
\(\Rightarrow S=\left(1+1+1+...+1\right)-\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{10000}\right)\)
\(\Rightarrow S=99-\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{10000}\right)< 99.\)
\(\Rightarrow S< 99\) (1).
Đặt \(A=\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{10000}\)
\(\Rightarrow A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)
Ta có:
\(\left\{{}\begin{matrix}\frac{1}{2^2}< \frac{1}{1.2}\\\frac{1}{3^2}< \frac{1}{2.3}\\....\\\frac{1}{100^2}< \frac{1}{99.100}\end{matrix}\right.\)
\(\Rightarrow A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(\Rightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow A< 1-\frac{1}{100}\)
Vì \(1-\frac{1}{100}< 1.\)
\(\Rightarrow A< 1.\)
\(\Rightarrow S>99-1\)
\(\Rightarrow S>98\) (2).
Từ (1) và (2) \(\Rightarrow98< S< 99.\)
\(\Rightarrow S\) không phải là số nguyên (đpcm).
Chúc bạn học tốt!
