Question

Write down and simplify :
The $8^{\text{th}}$ term of $(1+2x{)}^{-\frac{1}{2}}$

Answer 1

The general term i.e. $(r+1)th$ term in the expansion of $(1+x{)}^n$ is given by

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

Now,

For $T_8,$ $r=7$

$\therefore T_8=\frac{-\frac{1}{2}(-\frac{1}{2}-1)(-\frac{1}{2}-2)(-\frac{1}{2}-3)(-\frac{1}{2}-4)(-\frac{1}{2}-5)(-\frac{1}{2}-6)}{7!}(2x{)}^7$

$=\frac{(-1)(-3)(-5)(-7)(-9)(-11)(-13)}{2^7.7!}{2}^7.x^7$

$=-\frac{\mathrm{9.11.13}}{\mathrm{2.4.6}}{x}^7$

$=-\frac{429}{16}{x}^7$