cho a= 5/1.2.3+8/2.3.4+11/3.4.5+..+59/19.20.21 cmr a<2 28/09/2021 Bởi Lydia cho a= 5/1.2.3+8/2.3.4+11/3.4.5+..+59/19.20.21 cmr a<2
Giải thích các bước giải: Ta có: $A=\dfrac{5}{1.2.3}+\dfrac{8}{2.3.4}+\dfrac{11}{3.4.5}+…+\dfrac{59}{19.20.21}$ $\to A=\dfrac{1+2+3-1}{1.2.3}+\dfrac{2+3+4-1}{2.3.4}+\dfrac{3+4+5-1}{3.4.5}+…+\dfrac{19+20+21-1}{19.20.21}$ $\to A=\dfrac{1+2+3}{1.2.3}-\dfrac{1}{1.2.3}+\dfrac{2+3+4}{2.3.4}-\dfrac{1}{2.3.4}+\dfrac{3+4+5}{3.4.5}-\dfrac{1}{3.4.5}+…+\dfrac{19+20+21}{19.20.21}-\dfrac{1}{19.20.21}$ $\to A=\dfrac{1+2+3}{1.2.3}+\dfrac{2+3+4}{2.3.4}+\dfrac{3+4+5}{3.4.5}+…+\dfrac{19+20+21}{19.20.21}-(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+…+\dfrac{1}{19.20.21})$ $\to A=(\dfrac1{2.3}+\dfrac{1}{1.3}+\dfrac{1}{1.2})+(\dfrac{1}{3.4}+\dfrac{1}{2.4}+\dfrac{1}{2.3})+(\dfrac{1}{4.5}+\dfrac{1}{3.5}+\dfrac{1}{3.4})+…+(\dfrac{1}{20.21}+\dfrac{1}{19.21}+\dfrac{1}{19.20})-(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+…+\dfrac{1}{19.20.21})$ $\to A=(\dfrac{1}{2.3}+\dfrac{1}{3.4}+…+\dfrac{1}{20.21})+(\dfrac{1}{1.3}+\dfrac{1}{2.4}+…+\dfrac{1}{19.21})+(\dfrac{1}{1.2}+\dfrac{1}{2.3}+…+\dfrac{1}{19.20})-(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+…+\dfrac{1}{19.20.21})$ $\to A<(\dfrac{1}{2.3}+\dfrac{1}{3.4}+…+\dfrac{1}{20.21})+(\dfrac{1}{1.2}+\dfrac{1}{2.3}+…+\dfrac{1}{19.20})-(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+…+\dfrac{1}{19.20.21})$ $\to A<(\dfrac{3-2}{2.3}+\dfrac{4-3}{3.4}+…+\dfrac{21-20}{20.21})+(\dfrac{2-1}{1.2}+\dfrac{3-2}{2.3}+…+\dfrac{20-19}{19.20})-\dfrac12(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+\dfrac{2}{3.4.5}+…+\dfrac{2}{19.20.21})$ $\to A<(\dfrac12-\dfrac13+\dfrac13-\dfrac14+..+\dfrac1{20}-\dfrac1{21})+(\dfrac11-\dfrac12+\dfrac12-\dfrac13+…+\dfrac1{19}-\dfrac1{20})-\dfrac12(\dfrac{3-1}{1.2.3}+\dfrac{4-2}{2.3.4}+\dfrac{5-3}{3.4.5}+…+\dfrac{21-19}{19.20.21})$ $\to A<\dfrac{19}{42}+\dfrac{19}{20}-\dfrac12(\dfrac1{1.2}-\dfrac1{20.21})$ $\to A<\dfrac{19}{42}+\dfrac{19}{20}-\dfrac12(\dfrac1{2}-\dfrac1{20.21})$ $\to A<2$ Bình luận
Giải thích các bước giải:
Ta có:
$A=\dfrac{5}{1.2.3}+\dfrac{8}{2.3.4}+\dfrac{11}{3.4.5}+…+\dfrac{59}{19.20.21}$
$\to A=\dfrac{1+2+3-1}{1.2.3}+\dfrac{2+3+4-1}{2.3.4}+\dfrac{3+4+5-1}{3.4.5}+…+\dfrac{19+20+21-1}{19.20.21}$
$\to A=\dfrac{1+2+3}{1.2.3}-\dfrac{1}{1.2.3}+\dfrac{2+3+4}{2.3.4}-\dfrac{1}{2.3.4}+\dfrac{3+4+5}{3.4.5}-\dfrac{1}{3.4.5}+…+\dfrac{19+20+21}{19.20.21}-\dfrac{1}{19.20.21}$
$\to A=\dfrac{1+2+3}{1.2.3}+\dfrac{2+3+4}{2.3.4}+\dfrac{3+4+5}{3.4.5}+…+\dfrac{19+20+21}{19.20.21}-(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+…+\dfrac{1}{19.20.21})$
$\to A=(\dfrac1{2.3}+\dfrac{1}{1.3}+\dfrac{1}{1.2})+(\dfrac{1}{3.4}+\dfrac{1}{2.4}+\dfrac{1}{2.3})+(\dfrac{1}{4.5}+\dfrac{1}{3.5}+\dfrac{1}{3.4})+…+(\dfrac{1}{20.21}+\dfrac{1}{19.21}+\dfrac{1}{19.20})-(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+…+\dfrac{1}{19.20.21})$
$\to A=(\dfrac{1}{2.3}+\dfrac{1}{3.4}+…+\dfrac{1}{20.21})+(\dfrac{1}{1.3}+\dfrac{1}{2.4}+…+\dfrac{1}{19.21})+(\dfrac{1}{1.2}+\dfrac{1}{2.3}+…+\dfrac{1}{19.20})-(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+…+\dfrac{1}{19.20.21})$
$\to A<(\dfrac{1}{2.3}+\dfrac{1}{3.4}+…+\dfrac{1}{20.21})+(\dfrac{1}{1.2}+\dfrac{1}{2.3}+…+\dfrac{1}{19.20})-(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+…+\dfrac{1}{19.20.21})$
$\to A<(\dfrac{3-2}{2.3}+\dfrac{4-3}{3.4}+…+\dfrac{21-20}{20.21})+(\dfrac{2-1}{1.2}+\dfrac{3-2}{2.3}+…+\dfrac{20-19}{19.20})-\dfrac12(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+\dfrac{2}{3.4.5}+…+\dfrac{2}{19.20.21})$
$\to A<(\dfrac12-\dfrac13+\dfrac13-\dfrac14+..+\dfrac1{20}-\dfrac1{21})+(\dfrac11-\dfrac12+\dfrac12-\dfrac13+…+\dfrac1{19}-\dfrac1{20})-\dfrac12(\dfrac{3-1}{1.2.3}+\dfrac{4-2}{2.3.4}+\dfrac{5-3}{3.4.5}+…+\dfrac{21-19}{19.20.21})$
$\to A<\dfrac{19}{42}+\dfrac{19}{20}-\dfrac12(\dfrac1{1.2}-\dfrac1{20.21})$
$\to A<\dfrac{19}{42}+\dfrac{19}{20}-\dfrac12(\dfrac1{2}-\dfrac1{20.21})$
$\to A<2$