Question

Write down and simplify :
The ${10}^{\text{th}}$ term of $(1+3a^2{)}^{\frac{16}{3}}$

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 $T_{10},$ $r=9$

$\therefore T_{10}=\frac{\frac{16}{3}(\frac{16}{3}-1)(\frac{16}{3}-2)(\frac{16}{3}-3)(\frac{16}{3}-4)(\frac{16}{3}-5)(\frac{16}{3}-6)(\frac{16}{3}-7)(\frac{16}{3}-8)}{9!}(3a^2{)}^9$

$=\frac{\left(16\right)\left(13\right)\left(10\right)\left(7\right)\left(4\right)\left(1\right)(-2)(-5)(-8)}{3^9.9!}{\left(3\right)}^9.{\left(a^2\right)}^9$

$=-\frac{\mathrm{8.13.10}}{\mathrm{3.3.9}}{a}^{18}$

$=-\frac{1040}{81}{a}^{18}$