Question

7/4.( 33/12 + 3333/2020 + 333333/303030 + 33333333/42424242) = ?

Answer 1

\(\frac{7}{4}\left(\frac{33}{12}+\frac{3333}{2020}+\frac{333333}{303030}+\frac{33333333}{42424242}\right)\)

=\(\frac{7}{4}\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)\)

=\(\frac{7}{4}.33\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\)

=\(\frac{231}{4}\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)

=\(\frac{231}{4}\left(\frac{1}{3}-\frac{1}{7}\right)\)

=\(\frac{231}{4}.\frac{4}{21}\)

= 11