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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 )
\(3A=\frac{1}{6}-\frac{1}{8242206}=\frac{1373701}{8242206}-\frac{1}{8242206}=\frac{1373700}{8242206}\)
\(A=\frac{1373700}{8242206}:3=\frac{457900}{8242206}\)
\(\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+...+\frac{1}{27.28.29.30}\)
\(=\frac{1}{3}\left(\frac{3}{1.2.3.4}+\frac{3}{2.3.4.5}+...+\frac{3}{27.28.29.30}\right)\)
\(=\frac{1}{3}\left(\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+...+\frac{1}{27.28.29}-\frac{1}{28.29.30}\right)\)
\(=\frac{1}{3}\left(\frac{1}{1.2.3}-\frac{1}{28.29.30}\right)\)
Đặt \(A=\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+...+\frac{1}{17.18.19.20}\)
\(A=\frac{4-1}{1.2.3.4}+\frac{5-2}{2.3.4.5}+....+\frac{20-17}{17.18.19.20}\)
\(A=\frac{3}{1.2.3.4}+\frac{3}{2.3.4.5}+....+\frac{3}{17.18.19.20}\)
\(3A=\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+....+\frac{1}{17.18.19}-\frac{1}{18.19.20}\)
\(3A=\frac{1}{1.2.3}-\frac{1}{18.19.20}=\frac{1139}{6840}\)
\(\Rightarrow A=\frac{1139}{6840}\div3=\frac{1139}{20520}\)
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{98.99.100}-3x=\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+...+\frac{1}{27.28.29.30}\)
\(\Leftrightarrow\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)-3x=\frac{1}{3}.\left(\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+...+\frac{1}{27.28.29}-\frac{1}{28.29.30}\right)\)
\(\Leftrightarrow\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{99.100}\right)-3x=\frac{1}{3}.\left(\frac{1}{1.2.3}-\frac{1}{28.29.30}\right)\)
\(\Leftrightarrow\frac{4949}{19800}-3x=\frac{451}{8120}\)
\(\Leftrightarrow x\approx0,0648\)
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{98.99.100}-3x=\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+...+\frac{1}{27.28.29.30}\)
\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)-3x=\frac{1}{3}\left(\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+...+\frac{1}{27.28.29}-\frac{1}{28.29.30}\right)\)
\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)-3x=\frac{1}{3}\left(\frac{1}{1.2.3}-\frac{1}{28.29.30}\right)\)
\(\Rightarrow\frac{4949}{19800}-3x=\frac{451}{8120}\)
\(\Rightarrow3x=\frac{4949}{19800}-\frac{451}{8120}\)
\(\Rightarrow x=\left(\frac{4949}{19800}-\frac{451}{8120}\right):3\)
\(\Rightarrow\left(\frac{1}{1}-\frac{1}{30}\right)x=-3\)
\(\Rightarrow\frac{29}{30}x=-3\)
\(\Rightarrow x=\left(-\frac{29}{90}\right)\)
tính trog ngoặc trc nè :
\(\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+...+\frac{1}{27.28.29.30}\)
=\(\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}...+\frac{1}{27.28.29}-\frac{1}{28.29.30}\)
=\(\frac{1}{1.2.3}-\frac{1}{28.29.30}\)
=\(\frac{1}{6}-\frac{1}{24360}\)
=\(\frac{1353}{8120}\)
thay vô biểu thức :
\(\frac{1353}{8120}.x=-3\)
x=\(-\frac{8120}{451}\)
Đặt A = \(1+\frac{1}{1.2}+\frac{1}{1.2.3}+\frac{1}{1.2.3.4}+...+\frac{1}{1.2.3....n}\)
Ta có: \(\frac{1}{1.2}=\frac{1}{1.2}\)
\(\frac{1}{1.2.3}=\frac{1}{2.3}\)
\(\frac{1}{1.2.3.4}< \frac{1}{3.4}\)
..............
\(\frac{1}{1.2.3....n}< \frac{1}{\left(n-1\right)n}\)
Cộng vế với vế ta được:
\(A< 1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}=1+1-\frac{1}{n}=2-\frac{1}{n}< 2\)(đpcm)