Prime factors.

By Binomial expansion,

$(1+x)^{21} = ^{ 21 }{ { C } }_{0}{x}^{0} + ^{ 21}{ { C } }_{1}x + ^{ 21}{ { C } }_{2}{x}^{2} + ... + ^{ 21}{ { C } }_{20}{x}^{20} + ^{ 21}{ { C } }_{21}{x}^{21}$

Now differentiate both sides w.r.t $x$ , then

$21(1+x)^{20} = ^{ 21}{ { C } }_{1} + 2.^{ 21}{ { C } }_{2}x + ... + 20.^{ 21}{ { C } }_{20}{x}^{19} + 21.^{ 21}{ { C } }_{21}{x}^{20}$

Putting $x = +1, -1, i, -i$ (where $i$ is $\sqrt{-1}$ ) separately to form four equations and adding them we would get.

$21\times ({2}^{20} + 0 + {(1+i)}^{20} + {(1-i)}^{20}) = ^{ 21}{ { C } }_{1} + 5. ^{ 21}{ { C } }_{5} + 9. ^{ 21}{ { C } }_{9}$ $+ ... +$ $17. ^{ 21}{ { C } }_{17} + 21. ^{ 21}{ { C } }_{21} = k$

$\implies$ $k = 21\times ({2}^{20} - {2}^{11})$

$\implies$ $k = {2}^{11}\times 3 \times {7}^{2}\times 73$

Hence the required answer is $2\times 3\times 7\times 73 = 3066$ .

It is given that: $\displaystyle \sum _{k=0} ^5 {(4k+1)^{21}C_{4k+1}} = k$

$\displaystyle \sum _{k=0} ^5 {(4k+1)^{21}C_{4k+1}} = \sum _{k=0} ^5 {\frac{(4k+1)21!}{(4k+1)!(20-4k)!}} = \sum _{k=0} ^5 { \frac {21(20!)} {4k!(20-4k)!}}$

$\displaystyle = 21 \sum _{k=0} ^5 {^{20}C_{4k}} = 21(^{20} C _0 + ^{20} C _4 + ^{20} C _8 + ^{20} C _{12} + ^{20} C _{16} + ^{20} C _{20}$