Question

If $x$ is positive, the first negative term in the expansion of $(1+x{)}^{\frac{22}{5}}$ is

Answer 1

Since $x>0$

First negative term occurs when

$n=\frac{22}{5}$

$\frac{n(n-1)(n-2).....(n-r)}{1\times 2\times 3\times ......(r+1)}=-ve$

So, $\frac{22}{5}=4.2$

So, $n-5<0$

So, first negative tem

$\Rightarrow \frac{n(n-1)(n-2)(n-3)(n-4)(n-5)}{1\times 2\times 3\times 4\times 5\times 6}\times x^6\Rightarrow \frac{\frac{22}{5}\times \frac{17}{5}\times \frac{12}{5}\times \frac{7}{5}\times \frac{2}{5}\times \left(\frac{-3}{5}\right)}{1\times 2\times 3\times 4\times 5\times 6}\times x^6\Rightarrow \frac{-1309}{5^7}\times x^6=\frac{-1309}{78125}{x}^6$