Question

Write down and simplify :
The ${11}^{\text{th}}$ term of $(1-2x^3{)}^{-\frac{11}{2}}$

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 $T_{11},$ $r=10$

$\therefore T_{11}=\frac{-\frac{11}{2}(-\frac{11}{2}-1)(-\frac{11}{2}-2)(-\frac{11}{2}-3)(-\frac{11}{2}-4)(-\frac{11}{2}-5)(-\frac{11}{2}-6)(-\frac{11}{2}-7)(-\frac{11}{2}-8)(-\frac{11}{2}-9)}{10!}(-2x^3{)}^{10}$

$=\frac{(-11)(-13)(-15)(-17)(-19)(-21)(-23)(-25)(-27)(-29)}{2^{10}.10!}{(-2)}^{10}.{\left(x^3\right)}^{10}$

$=\frac{\mathrm{11.13.15.17.19.21.23.25.27.29}}{\mathrm{1.2.3.4.5.6.7.8.9.10}}{x}^{30}$

$=\frac{46189\times 10005}{256}{x}^{30}$