Tính tổng sau bằng cách hợp lí
A = \(\dfrac{1}{4}+\dfrac{1}{12}+\dfrac{1}{36}+...+\dfrac{1}{972}+\dfrac{1}{2916}\)
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\(B=\dfrac{1}{4}+\dfrac{1}{12}+\dfrac{1}{36}+\dfrac{1}{108}+\dfrac{1}{324}+\dfrac{1}{972}\\\)
\(3B=3\left(\dfrac{1}{4}+\dfrac{1}{12}+\dfrac{1}{36}+\dfrac{1}{108}+\dfrac{1}{324}+\dfrac{1}{972}\right)\)
\(3B=\dfrac{3}{4}+\dfrac{3}{12}+\dfrac{3}{36}+\dfrac{3}{108}+\dfrac{3}{324}+\dfrac{3}{972}\)
\(3B=\dfrac{3}{4}+\dfrac{1}{4}+\dfrac{1}{12}+\dfrac{1}{36}+\dfrac{1}{108}+\dfrac{1}{324}\)
\(2B=3B-B\)
\(2B=\left(\dfrac{3}{4}+\dfrac{1}{4}+\dfrac{1}{12}+\dfrac{1}{36}+\dfrac{1}{108}+\dfrac{1}{324}\right)-\left(\dfrac{1}{4}+\dfrac{1}{12}+\dfrac{1}{36}+\dfrac{1}{108}+\dfrac{1}{324}+\dfrac{1}{972}\right)\)
\(2B=\dfrac{3}{4}-\dfrac{1}{972}=\dfrac{729-1}{972}=\dfrac{728}{972}=\dfrac{182}{243}\)
\(B=\dfrac{182}{243}:\dfrac{1}{2}=\dfrac{182\cdot2}{243}=\dfrac{364}{243}\)
\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2021.2022}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2021}-\dfrac{1}{2022}\)
\(=1-\dfrac{1}{2022}=\dfrac{2021}{2022}\)
\(B=\dfrac{4}{3.7}+\dfrac{4}{7.11}+\dfrac{4}{11.15}+...+\dfrac{4}{107.111}\)
\(=\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{15}+...+\dfrac{1}{107}-\dfrac{1}{111}\)
\(=\dfrac{1}{3}-\dfrac{1}{111}=\dfrac{12}{37}\)
\(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}\\ =\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}\\ =\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}\\ =\dfrac{1}{3}-\dfrac{1}{9}\\ =\dfrac{2}{9}\)
A=14+112+136+...+1972+12916
3A=34+14+112+...+1324+1972
3A−A=(34+14+112+...+1324+1972)−(14+112+136+...+1972+12916)
2A=34−12916
A=10932916