Question

(1+1/2).(1.1/3)x ...x(1+1/2019)

Answer 1

\(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)..\left(1+\frac{1}{2019}\right)\)

\(=\frac{3}{2}\cdot\frac{4}{3}\cdot...\cdot\frac{2020}{2019}\)

\(=\frac{2020}{2}=1010\)

Answer 2

\(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)...\left(1+\frac{1}{2019}\right)\)

\(=\frac{3}{2}\cdot\frac{4}{3}\cdot...\cdot\frac{2020}{2019}\)

\(=\frac{2020}{2}\)

\(=1010\)

Answer 3

\(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)...\left(1+\frac{1}{2019}\right)\)

\(=\frac{3}{2}.\frac{4}{3}....\frac{2020}{2019}\)

\(=\frac{2020}{2}\)

\(=1010\)