\(C=\frac{4-1}{4}+\frac{9-1}{9}+....+\frac{10000-1}{10000}.\)
\(C=1-\frac{1}{4}+1-\frac{1}{9}+.....+1-\frac{1}{10000}.\)
\(C=\left(1+1+....+1\right)-\left(\frac{1}{4}+\frac{1}{9}+...+\frac{1}{10000}\right)\)
\(C=99-\left(\frac{1}{4}+\frac{1}{9}+...+\frac{1}{10000}\right)\)
ta có :\(\frac{1}{4}< 1,\frac{1}{9}< 1,......,\frac{1}{10000}< 1\)
\(\Rightarrow\frac{1}{4}+\frac{1}{9}+...+\frac{1}{10000}< 1\)
\(C=99-\left(\frac{1}{4}+\frac{1}{9}+...+\frac{1}{10000}\right)>98\)
vậy C>98
\(C=\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{9999}{10000}\)
\(=\left(1-\frac{1}{4}\right)+\left(1-\frac{1}{9}\right)+\left(1-\frac{1}{16}\right)+...+\left(1-\frac{1}{10000}\right)\)
\(=\left(1+1+1+...+1\right)-\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{10000}\right)\)
\(=99-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\right)\)
Đặt D = \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)
Ta có: \(\frac{1}{2^2}< \frac{1}{1.2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}\)
\(\frac{1}{4^2}< \frac{1}{3.4}\)
.............
\(\frac{1}{100^2}< \frac{1}{99.100}\)
\(\Rightarrow D>\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+..+\frac{1}{99.100}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}< 1\)
\(\Rightarrow C=99-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\right)>98\)(đpcm)
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