JEE Main 2019 — Binomial Theorem Question with Solution

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

We have $\frac{{ }^{20}{C}_{i-1}}{{ }^{20}{C}_i+{ }^{20}{C}_{i-1}}$

Using ${}^{n}C_{r-1}+{}^{n}C_r={}^{n+1}C_r,$ we get

$\frac{{ }^{20}{C}_{i-1}}{{ }^{20}{C}_i+{ }^{20}{C}_{i-1}}=\frac{{ }^{20}{C}_{i-1}}{{ }^{21}{C}_i}$

Now, using ${}^{n}C_r=\frac{n}{r}\times {}^{n-1}C_{r-1},$ we get

$\frac{{ }^{20}{C}_{i-1}}{{ }^{21}{C}_i}=\frac{{}^{20}{C}_{i-1}}{{\displaystyle \frac{21}{i}}\times \left({}^{20}{C}_{i-1}\right)}=\frac{i}{21}$

Hence, the given sum is 

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

Now, using $\underset{r=1}{\overset{n}{\sum r^3}}={\left(\frac{n\left(n+1\right)}{2}\right)}^2,$ we get

$\frac{1}{{\left(21\right)}^3}{\sum }_{i=1}^{20}{i}^3=\frac{1}{{\left(21\right)}^3}{\left(\frac{20·21}{2}\right)}^2=\frac{100}{21}$

Using the given sum, we get $\frac{100}{21}=\frac{k}{21}$

$\Rightarrow k=100.$