Question

Tính

P= 1/1*2*3 + 1/2*3*4 + 1/3*4*5 +........+1/98*99*100

Answer 1

\(P=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{98.99.100}\)

\(P=\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+...+\frac{100-98}{98.99.100}\)

\(\Rightarrow2P=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\)

\(\Rightarrow2P=\frac{1}{1.2}-\frac{1}{99.100}\)

\(\Rightarrow P=\left(\frac{1}{2}-\frac{1}{9900}\right)\div2\)

\(\Rightarrow P=\frac{4949}{9900}\cdot\frac{1}{2}=\frac{4949}{19800}\)

Answer 2

\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}\)

\(2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{98.99.100}\)

\(2A=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\)

\(2A=\frac{1}{1.2}-\frac{1}{99.100}\)

\(2A=\frac{4949}{9900}\)

\(A=\frac{4949}{19800}\)