Tính 20.21+22.23+23.24+...+39.40
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đặt S=20.21+22.23+23.24+....+38.39+39.40
3s = 3.20.21+3.22.23+3.23.24+...+3.38.39+3.39.40
=20.21.(22-19)+22.23.(24-21)+23.24.(25-22)+...+39.40.(41-38)
=39.40.41-19.20.21=55980
suy ra s = 18660
\(S=20.21+22.23+23.24+....+39.40\)
\(3S=3.20.21+3.22.23+3.23.24+...+3.39.40\)
\(3S=20.21.\left(22-19\right)+22.23.\left(24-21\right)+....+39.40.\left(41-38\right)\)
\(3S=39.40.41-19.20.21\)
\(3S=55980\)
\(S=18660\).
Ta có: \(S=20.21+21.22+22.23+....+39.40\)
\(\Rightarrow3S=3.20.21+3.21.22+.....+3.39.40\)
\(\Rightarrow3S=20.21.\left(22-19\right)+22.23.\left(24-21\right)+.....+39.40.\left(41-38\right)\)
\(\Rightarrow3S=39.40.41-19.20.21\)
\(\Rightarrow3S=55980\)
\(\Rightarrow S=18660\)
Ta có:
\(\frac{1}{20.21}+\frac{1}{21.22}+\frac{1}{22.23}+...+\frac{1}{60.61}\)
\(=\frac{1}{20}-\frac{1}{21}+\frac{1}{21}-\frac{1}{22}+\frac{1}{22}-\frac{1}{23}+...+\frac{1}{60}-\frac{1}{61}\)
\(=\frac{1}{2}-\frac{1}{61}=\frac{59}{122}\)
b) \(\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+...+\frac{4}{45.49}\)
\(=\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+...+\frac{1}{45.49}\)
\(=\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+...+\frac{1}{45}-\frac{1}{49}\)
\(=\frac{1}{5}-\frac{1}{49}=\frac{44}{245}\)
Bn Tấn sai rùi
phần a , câu cuối là \(\frac{1}{20}\)chứ đâu phải \(\frac{1}{2}\)