Question

Expand the following:
${(y^2+\frac{2}{y})}^4,(y\ne 0)$

Answer 1

${(y^2+\frac{2}{y})}^4$

As we know that the expansion of the expression ${(a+b)}^n$ will be as follows-

${(a+b)}^n={}_n{C}_0{a}^n{b}^0+{}_n{C}_1{a}^{n-1}{b}^1+.............+{}_n{C}_n{a}^0{b}^n$

Therefore, in the given expansion ${(y^2+\frac{2}{y})}^4$,

$a=y^2$

$b=\frac{2}{y}$

$n=4$

$\therefore$ Expansion of the given expression-

${(y^2+\frac{2}{y})}^4={}_4{C}_0{\left(y^2\right)}^4{\left(\frac{2}{y}\right)}^0+{}_4{C}_1{\left(y^2\right)}^3{\left(\frac{2}{y}\right)}^1+{}_4{C}_2{\left(y^2\right)}^2{\left(\frac{2}{y}\right)}^2+{}_4{C}_3{\left(y^2\right)}^1{\left(\frac{2}{y}\right)}^3+{}_4{C}_4{\left(y^2\right)}^0{\left(\frac{2}{y}\right)}^4$

$\because {}_n{C}_r=\frac{n!}{r!(n-r)!}$

$\therefore {(y^2+\frac{2}{y})}^4=1.y^8+4.y^6\left(\frac{2}{y}\right)+6.y^4\left(\frac{4}{y^2}\right)+4.y^2\left(\frac{8}{y^3}\right)+1.y^0\left(\frac{16}{y^4}\right)$

$\Rightarrow {(y^2+\frac{2}{y})}^4=y^8+8y^5+24y^2+\frac{32}{y}+\frac{16}{y^4}$

Hence, ${(y^2+\frac{2}{y})}^4=y^8+8y^5+24y^2+\frac{32}{y}+\frac{16}{y^4}$.