=1/1.3-1/3.5+1/3.5-1/5.7+...+1/99.11-1/11.13
=1/1.3-1/11.13
=1/3-1/143
=140/429
=1/1.3-1/3.5+1/3.5-1/5.7+...+1/99.11-1/11.13
=1/1.3-1/11.13
=1/3-1/143
=140/429
\(\frac{4}{1.3.5}+\frac{4}{3.5.7}+\frac{4}{5.7.9}\)
\(A:\)\(\frac{4}{9.11.13}+\frac{4}{11.13.15}+...+\frac{4}{59.61.63}\)
\(B:\) \(\frac{1}{1.3.5}+\frac{1}{3.5.7}+...+\frac{1}{47.49.51}\)
\(B=\frac{1}{1.3.5}+\frac{1}{3.5.7}+...+\frac{1}{47.49.51}\)
\(C=\frac{4}{9.11.13}+\frac{4}{11.13.15}+...+\frac{4}{59.61.63}\)
Ai nhanh mik tick nha !!!!!
tính nhanh
\(\frac{3}{2.6}+\frac{3}{6.10}+\frac{3}{10.14}\)
\(\frac{4}{1.3.5}+\frac{4}{3.5.7}+\frac{4}{5.7.9}\)
\(\frac{5}{2.4.6}+\frac{5}{4.6.8}+\frac{5}{6.8.10}\)
A = \(\frac{2.4.6+4.6.8+6.8.10+8.10.12+...+198.200.202}{1.3.5+3.5.7+5.7.9+7.9.11+...+97.99.101}\) =?
Chứng minh rằng :
a) A=\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{18.19.20}<\frac{1}{4}\)
b) B=\(\frac{36}{1.3.5}+\frac{36}{3.5.7}+\frac{36}{5.7.9}+...+\frac{36}{25.27.29}<3\)
Tính :C=\(\frac{3}{3.5.7}+\frac{3}{5.7.9}+...+\frac{3}{2001.2003.2005}\)
thu gọn tổng :
\(B=\frac{1}{1.3.5}+\frac{1}{3.5.7}+\frac{1}{5.7.9}+...+\frac{1}{2017.2019.2021}\)
chứng minh rằng
\(B=\frac{36}{1.3.5}+\frac{36}{3.5.7}+\frac{36}{5.7.9}+...+\frac{36}{25.27.29}\)