Question

$\sum _{i=1}^{20}\left(\frac{{}^{20}{C}_{i-1}}{{}^{20}{C}_i+{}^{20}{C}_{i-1}}\right)=\frac{k}{21}$, then k equals?

• A. $200$
• B. $50$
• C. $100$
• D. $400$

Answer 1

The correct option is C $100$

$\sum _{i=1}^{20}{\left(\frac{{}^{20}{C}_{i-1}}{{}^{20}{C}_i+{}^{20}{C}_{i-1}}\right)}^3=\frac{k}{21}$

$\Rightarrow \sum _{i=1}^{20}{\left(\frac{{}^{20}{C}_{i-1}}{{}^{21}{C}_i}\right)}^3=\frac{k}{21}$

$\Rightarrow \sum _{i=1}^{20}{\left(\frac{i}{21}\right)}^3=\frac{k}{21}$

$\Rightarrow \frac{1}{(21{)}^3}{\left[\frac{20\left(21\right)}{2}\right]}^2=\frac{k}{21}$

$\Rightarrow 100=k$.