Question

Find the $5^{th}$ term in the expansion of ${\left(2x+\frac{1}{3y}\right)}^8.$

Answer 1

$T_{r+1}{=}^n{C}_r{x}^{n-r}{a}^r$

where $x\to 2x,a\to \frac{1}{3y}$ and $n=8$

$\therefore T_{4+1}{=}^8{C}_4{\left(2x\right)}^{8-4}{\left(\frac{1}{3y}\right)}^4$

${=}^8{C}_4{\left(2x\right)}^4{\left(\frac{1}{3y}\right)}^4$

$=\frac{8!}{4!4!}{2}^4{x}^4\frac{1}{3^4}\times \frac{1}{y^4}$

$=\frac{8\times 7\times 6\times 5}{4\times 3\times 2}\times 16\times \frac{1}{81}\times \frac{x^4}{y^4}$

$=\frac{7\times 2\times 5\times 16}{81}\times \frac{x^4}{y^4}$

$=\frac{1120x^4}{81y^4}$