Question

Write down and simplify :
The $7^{\text{th}}$ term of $(3^8+6^{4}x{)}^{\frac{11}{4}}$

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,

$(3^8+6^{4}x{)}^{\frac{11}{4}}=3^{22}{(1+\frac{6^4}{3^8}x)}^{\frac{11}{4}}=3^{22}{(1+\frac{2^4}{3^4}x)}^{\frac{11}{4}}$

For $T_7,$ $r=6$

$\therefore T_7=3^{22}[\frac{\frac{11}{4}(\frac{11}{4}-1)(\frac{11}{4}-2)(\frac{11}{4}-3)(\frac{11}{4}-4)(\frac{11}{4}-5)}{6!}{\left(\frac{2^4}{3^4}\right)}^6.x^6]$

$=3^{22}[\frac{\left(11\right)\left(7\right)\left(3\right)(-1)(-5)(-9)}{4^6{.3}^{24}.6!}{2}^{24}.x^6]$

$=-3^{22}[\frac{\mathrm{11.7.1}}{3^{23}}{2}^8.x^6]$

$=-\frac{19712}{3}{x}^6$