Question

Find the first negative term in the expansion of ${(1+x)}^{27/5}$ where x is a positive number.

Answer 1

$(1+x{)}^k=1+kx+\frac{k(k-1)}{2!}{x}^2+.....+\frac{k(k-1)(k-2)....(k-n+1)}{n!}{x}^n$

$(1+x{)}^{\frac{27}{5}}=1+\frac{27}{5}x+\frac{27}{5}\times \frac{22}{5}\times \frac{1}{2}{x}^2+.....$

First negative term will be when you have $\frac{27}{5}-n+1<0$ or $n>\frac{32}{5}$

The first term that fits this is the $7th$ term.

$\frac{\frac{27}{5}\times \frac{22}{5}\times \frac{17}{5}\times \frac{12}{5}\times \frac{7}{5}\times \frac{2}{5}\times \frac{-3}{5}}{7!}$ $x^7$