Question

the term independent of x in the expansion of [4x³+7/x²]¹⁴​

Answer 1

$\large\underline{\sf{Solution-}}$

↝Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

$\sf \: The \: general \: term \: of \: an \: expansion \: of \: {(x + y)}^{n} \: is \:$

$\underline{ \boxed{ \bf \: T_{(r+1)} = \:^n C_r \: {x}^{(n - r)} {y}^{r} }}$

↝ Now,

$\rm :\longmapsto\:In \: expansion \: of \: {\bigg( {4x}^{3} + \dfrac{7}{ {x}^{2} } \bigg) }^{14}$

$\rm :\longmapsto\:The \: general \: term \: is \: given \: by$

$\rm :\longmapsto\:T_{(r+1)} = \:^{14} C_r \: {\bigg( {4x}^{3} \bigg) }^{14 - r} {\bigg(\dfrac{7}{ {x}^{2} } \bigg) }^{r}$

$\rm :\longmapsto\:T_{(r+1)} = \:^{14} C_r \: {4}^{(14 - r)} {(x)}^{42 - 3r} \dfrac{ {7}^{r} }{ {x}^{2r} }$

$\rm :\longmapsto\:T_{(r+1)} = \:^{14} C_r \: {(4)}^{14 - r} {(7)}^{r} {(x)}^{42 - 3r - 2r}$

$\rm :\longmapsto\:T_{(r+1)} = \:^{14} C_r \: {(4)}^{14 - r} {(7)}^{r} {(x)}^{42 - 5r}$

↝ Now, for term to be independent of 'x',

$\rm :\longmapsto\:Put \: 42 - 5r = 0$

$\rm :\longmapsto\:5r = 42$

$\bf\implies \:r = \dfrac{42}{5} = 8.4 \: \: \cancel\in \: integer.$

$\bf\implies \:There \: is \: no \: term \: in \: expansion \: independent \: of \: x$

Additional Information :-

$\underline{ \boxed{ \bf \: If \: \:^n C_x = \:^n C_y \: \implies \: x = y \: \: or \: \: n = x + y}}$

$\underline{ \boxed{ \bf \: \:^n C_r + \:^n C_{r - 1} = \:^{n + 1} C_r}}$

$\underline{ \boxed{ \bf \: \dfrac{\:^n C_r}{\:^n C_{r - 1}} = \dfrac{n - r + 1}{r}}}$