Question

 C= (1-1/1+2)(1-1/1+2+3)(1-1/1+2+3+4).....(1-1/1+2+3+....+2016)

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Answer 1

\(C=\left(1-\dfrac{1}{1+2}\right)\left(1-\dfrac{1}{1+2+3}\right)\left(1-\dfrac{1}{1+2+3+4}\right)...\left(1-\dfrac{1}{1+2+3+...+2016}\right)\)

\(=\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{6}\right)...\left(1-\dfrac{1}{\dfrac{\left(2016+1\right).2016}{2}}\right)\)

\(=\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{6}\right)...\left(1-\dfrac{1}{2033136}\right)\)

\(=\dfrac{2}{3}.\dfrac{5}{6}...\dfrac{2033135}{2033136}\)

\(=\dfrac{4}{6}.\dfrac{10}{12}...\dfrac{4066270}{4066272}\)

\(=\left(\dfrac{1}{2}.\dfrac{2}{3}...\dfrac{2015}{2016}\right).\left(\dfrac{4}{3}.\dfrac{5}{4}...\dfrac{2018}{2017}\right)\)

\(=\dfrac{1}{2016}.\dfrac{2018}{3}=\dfrac{1009}{3024}\)