Question

Find the 8th term of ${\left(1-\frac{5x}{2}\right)}^{-3/5}$

Answer 1

${(1-\frac{5x}{2})}^{-\frac{3}{5}}$

General term of the expansion ${(1+X)}^n$ is given as-

$T_{r+1}=\frac{n(n-1)(n-2).....(n-r+1)}{r!}{X}^r$

Now, in the given expansion, i.e.,

${(1-\frac{5x}{2})}^{-\frac{3}{5}}$

$X=-\frac{5x}{2}$

$n=-\frac{3}{5}$

Therefore,

$T_8=T_{7+1}$

$\therefore r=7$

$T_8=\frac{(-\frac{3}{5})(-\frac{3}{5}-1)(-\frac{3}{5}-2).....(-\frac{3}{5}-7+1)}{7!}{(-\frac{5x}{2})}^7$

$\Rightarrow T_8=-{\left(\frac{5}{2}\right)}^7\frac{(-3)(-8)(-13).....(-33)}{5^{7}7!}{x}^7$

$\Rightarrow T_8=\frac{3\cdot 8\cdot 13\cdot .....\cdot 33}{2^{7}7!}{x}^7$

Hence the $8^{th}$ term of the given expansion is $\frac{3\cdot 8\cdot 13\cdot .....\cdot 33}{2^{7}7!}{x}^7$.