Question

Cm B=\(\frac{3}{1^2\cdot2^2}\)+\(\frac{5}{2^2\cdot3^2}\)+\(\frac{7}{3^2\cdot4^2}\)+...+\(\frac{19}{9^2\cdot10^2}\)<1

Answer 1

Ta có:(ĐỀ)=3/1.4+5/4.9+7/9.16+.....+19/81.100

                 =1/1.4 +1/4.9 +1/9.16+....+1/81.100

                =1-1/4+1/4-1/9+1/9-1/16+.....+1/81-1/100

                =1-1/100<1  =>B<1

MK ĐẦU TIÊN NHA BẠN!

Answer 2

\(B=\frac{3^2}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+.....+\frac{19}{9^2.10^2}\)

\(B=\frac{2^2-1^2}{1^2.2^2}+\frac{3^2-2^2}{2^2.3^2}+\frac{4^2-3^2}{3^2.4^2}+.....+\frac{10^2-9^2}{9^2.10^2}\)

\(B=\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+\frac{1}{3^2}-\frac{1}{4^2}+...+\frac{1}{9^2}-\frac{1}{10^2}\)

\(B=\frac{1}{1^2}-\frac{1}{10^2}=1-\frac{1}{10^2}<1\left(đpcm\right)\)