The first negative coefficient in the terms occurring in the expansion of $(1+x{)}^{\frac{21}{5}}$ is

The expansion of  ${\left(1+x\right)}^{\frac{21}{5}}=1+\frac{21}{5}x+\frac{\frac{21}{5}\left(\frac{21}{5}-1\right)}{2!}{x}^2+\frac{\frac{21}{5}\left(\frac{21}{5}-1\right)\left(\frac{21}{5}-2\right)}{3!}{x}^3+\dots$

Simplify the above expansion.

${\left(1+x\right)}^{\frac{21}{5}}=1+\frac{21}{5}x+\frac{21\times 16}{5^2\times 2!}{x}^2+\frac{21\times 16\times 11}{5^3\times 3!}{x}^3+\frac{21\times 16\times 11\times 6}{5^4\times 4!}+..$

The first negative coefficient in the terms occurring in the expansion of $(1+x{)}^{\frac{21}{5}}$ is

$=-\frac{21\times 16\times 11\times 6\times 1\times 4}{5^6\times 6!}$

$=\frac{-88704}{5^7\times 144}$

$=\frac{-616}{5^7}$