Find the coefficient of \(x^2\) in the binomial expansion of \(\left(x+\frac{2}{x^2}\right)^5\)

Answer Details

(x+2x2)5 $<br>$

n = 5,  r = 4,  p = x  and q = 2x2 $\frac{\mathrm{<br>}}{{\mathrm{<span\; style="font-weight:\; var(--bs-body-font-weight);"><br></span>}}^{}}$

5C4 ${}_4$x4 ${}^4$ (2x2 $\frac{\mathrm{<br>}}{}$)1 = 5C4 ${}_4$ 2x4x2 $\frac{\mathrm{<br>}}{}$

5C4 ${}_4$ 2x2 ${}^2$ = 5![5−4]!4! $\frac{\mathrm{<br>}}{}$ * 2x2 ${}^{\mathrm{<br>}}$

5∗4!4!∗2x2 $\frac{5\ast 4!}{4!}\ast 2{\mathrm{<br>}}^{}$ = 5 * 2x2 ${}^2$ = 10x2 ${}^{\mathrm{<br>}}$

The coefficient is 10.