Coefficient of the term independent of $x$ in the expansion $\left( x + \frac{1}{x^{2}} \right)^{6}\ is\ equal\ to$

Question

Dr. Abubacker · Accepted Answer

(b)
$General\ term,\ T_{r + 1} = \ ^{6}C_{r}x^{6 - r}\left( \frac{1}{x^{2}} ight)^{r}$
$\Longrightarrow T_{r + 1} = \ ^{6}C_{r}x^{6 - 3r}$
For term independent of $x,$ put
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 6 - 3r = 0$
$\Longrightarrow r = 2$
$	herefore$ Coefficient of independent term $= \ ^{6}C_{2} = 15$

Q. Coefficient of the term independent of $x$ in the expansion $\left( x + \frac{1}{x^{2}} \right)^{6}\ is\ equal\ to$

Explanation

Reviewed by Dr. Abubacker

Want more practice?