Tính
\(A=\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+\frac{1}{3.4.5.6}+...+\frac{1}{47.48.49.50}\)
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F= \(\frac{1}{1.2.3}\)- \(\frac{1}{2.3.4}\)+ \(\frac{1}{2.3.4}\)- \(\frac{1}{3.4.5}\)+....+\(\frac{1}{47.48.49}\)- \(\frac{1}{48.49.50}\)
F=\(\frac{1}{1.2.3}\)- \(\frac{1}{48.49.50}\)
F=\(\frac{6533}{39200}\)
\(\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+...+\frac{1}{47.48.49.50}\)
\(=\frac{1}{3}\cdot\left(\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+...+\frac{1}{47.48.49}-\frac{1}{48.49.50}\right)\)
\(=\frac{1}{3}\cdot\left(\frac{1}{1.2.3}-\frac{1}{48.49.50}\right)\)
\(=\frac{1}{3}\cdot\frac{6533}{39200}=\frac{6533}{117600}\)
\(\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+...+\frac{1}{97.98.99.100}=\frac{1}{3}.\left(\frac{3}{1.2.3.4}+\frac{3}{2.3.4.5}+...+\frac{3}{97.98.99.100}\right)=\frac{1}{3}.\left(\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+...+\frac{1}{97.98.99}-\frac{1}{98.99.100}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{1.2.3}-\frac{1}{98.99.100}\right)=\frac{1}{3}.\left(\frac{1}{6}-\frac{1}{970200}\right)=\frac{1}{18}-\frac{1}{6.970200}\)
\(\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+...+\frac{1}{97.98.99.100}\)
\(=\frac{1}{3}.\left(\frac{3}{1.2.3.4}+ \frac{3}{2.3.4.5}+...+\frac{3}{97.98.99.100}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+...+\frac{1}{97.98.99}-\frac{1}{98.99.100}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{1.2.3}-\frac{1}{98.99.100}\right)\)
\(=\frac{1}{3}.\frac{161699}{970200}=\frac{161699}{299106000}\)
Lại phải giải hết
Gọi dãy số trên là A
\(A=\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+.....+\frac{1}{200.201.202.203}\)
\(3A=\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-.....+\frac{1}{200.201.202}-\frac{1}{201.202.203}\)
\(3A=\frac{1}{1.2.3}-\frac{1}{201.202.203}\)(chỗ này lm hơi tắt tí )
\(3A=\frac{1}{6}-\frac{1}{8242206}=\frac{1373701}{8242206}-\frac{1}{8242206}=\frac{1373700}{8242206}\)
\(A=\frac{1373700}{8242206}:3=\frac{457900}{8242206}\)