\(\left(1+\dfrac{1}{3}\right)\left(1+\dfrac{1}{8}\right)\left(1+\dfrac{1}{15}\right)...\left(1+\dfrac{1}{120}\right)\)
= \(\dfrac{4}{3}.\dfrac{9}{8}.\dfrac{16}{15}.....\dfrac{121}{120}\)
= \(\dfrac{2^2}{1.3}+\dfrac{3^2}{2.4}.\dfrac{4^2}{3.5}.....\dfrac{11^2}{10.12}\)
= \(\dfrac{2}{1}.\dfrac{2}{3}.\dfrac{3}{2}.\dfrac{3}{4}.\dfrac{4}{3}.\dfrac{4}{5}.....\dfrac{11}{10}.\dfrac{11}{12}\)
= \(\dfrac{2}{1}\left(\dfrac{2}{3}.\dfrac{3}{2}\right)\left(\dfrac{3}{4}.\dfrac{4}{3}\right)...\left(\dfrac{10}{11}.\dfrac{11}{10}\right).\dfrac{11}{12}\)
= \(2.\dfrac{11}{12}\)
= \(\dfrac{11}{6}\)
\(\left(1+\frac{1}{3}\right)\left(1+\frac{1}{8}\right)\left(1+\frac{1}{15}\right)....\left(1+\frac{1}{120}\right)\\ =\frac{4}{3}.\frac{9}{8}.\frac{16}{15}...\frac{121}{120}\\ =\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}....\frac{11^2}{10.12}\\ \)
\(=\frac{2.11}{1.12}=\frac{11}{6}\)
nếu nó ở cuối có phép tính thì sẽ ra , nhg ở cuối k có phép mik cũng chịu , ở trên đây tôi giải nó k như vậy đâu .
!!!!
Help me!!
