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\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}\)
\(=\frac{3}{1.4}+\frac{5}{4.9}+\frac{7}{9.16}+...+\frac{19}{81.100}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+...+\frac{1}{81}-\frac{1}{100}\)
\(=1-\frac{1}{100}< 1\)
=> đpcm
Ủng hộ mk nha ^_-
\(B=\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+...+\frac{19}{9^2.10^2}\)
\(B=\frac{3}{1.4}+\frac{5}{4.9}+...+\frac{19}{81.100}\)
\(B=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+...+\frac{1}{81}-\frac{1}{100}\)
\(B=1-\frac{1}{100}< 1\left(đpcm\right)\)
\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+...+\frac{19}{9^2.10^2}=\frac{3}{1.4}+\frac{5}{4.9}+...+\frac{19}{81.100}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+...+\frac{1}{81}-\frac{1}{100}=1-\frac{1}{100}<1\)
Vậy \(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+...+\frac{19}{9^2.10^2}<1\)
\(B=\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+....+\frac{19}{9^2.10^2}\)
\(B=\frac{3}{1.4}+\frac{5}{4.9}+\frac{7}{9.16}+....+\frac{19}{81.100}\)
\(B=\frac{4-1}{1.4}+\frac{9-4}{4.9}+\frac{16-9}{9.16}+....+\frac{100-81}{81.100}\)
\(B=\frac{4}{1.4}-\frac{1}{1.4}+\frac{9}{4.9}-\frac{4}{4.9}+\frac{16}{9.16}-\frac{9}{9.16}+...+\frac{100}{81.100}-\frac{81}{81.100}\)
\(B=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+....+\frac{1}{81}-\frac{1}{100}\)
\(B=1-\frac{1}{100}< 1\)
=> B < 1 (Đpcm)
B = 3/12.22 + 5/22.32 + 7/32.42 + ... + 19/92.102
B = 3/1.4 + 5.4.9 + 7/9.16 + ... + 19/81.100
B = 1 - 1/4 + 1/4 - 1/9 + 1/9 - 1/16 + ... + 1/81 - 1/100
B = 1 - 1/100 < 1 ( đpcm)