Question

How to find coefficient term of independent of x of the expansion (x^1/12+x^-1/9)^14?

Answer 1

❏ QuesTion:-

$\sf \red{ find \: the \: coefficient \: of \: term \: independent} \\ \sf \green{of \: x \: in \: the \: expansion} \: of \: { \big({x}^{ \frac{1}{12} } + {x}^{ \frac{ - 1}{9} } \big) }^{14}$

❏ Answer:-

$\to \sf \: 3003 \: \: or \: \: \: ^{14}C_{6} \:$

❏ To Find :-

→ Term independent of x

❏ Explanation :-

We have ,

$\to \sf \: { \big({x}^{ \frac{1}{12} } + {x}^{ \frac{ - 1}{9} } \big) }^{14} \\ \\ \sf term \: independent \: of \: x \: means \: the \: term \: \\ \sf which \: is \: not \: depend \: on \: x \: ( {x}^{0}) \\ \bf \: hence \\ \\ \because \sf \: for \: coefficient \: of \: x \: \: \{ ^{n}C_{r} \: {a}^{n - r} {x}^{r} \} \\ \\ \to \sf ^{14}C_{r} \: {( {x}^{ \frac{1}{12} } )}^{14 - r} { {(x)}^{ \frac{ - 1}{9} } }^{r} \\ \\ \to \: \sf ^{14}C_{r} \: {( {x}^{ } )}^{ (\frac{14 - r}{12} )} { {(x)}^{ (\frac{ - r}{9}) } }^{} \: \\ \\ \to \: \sf ^{14}C_{r} \: {(x)}^{( \frac{14 - r}{12} - \frac{r}{9} )} \\ \\ \sf for \: term \: independent \: of \: x \\ \\ \to \sf \frac{14 - r}{12} - \frac{r}{9} = 0 \\ \\ \to \sf \frac{14 - r}{12} = \frac{r}{9} \\ \\ \to \sf 12r = 126 - 9r \\ \\ \to \sf \: 21r = 126 \\ \\ \to \boxed{\sf r = 6} \\ \\ \sf \: hence \: coefficient \: of \: {x}^{0} \: is \: - \\ \\ \to \: \sf ^{14}C_{6} \: {x}^{0} \\ \\ \to \: ^{14}C_{6} \: \\ \\ \to \: \frac{14 \times 13 \times 12 \times 11 \times 10 \times 9 }{6 \times 5 \times 4 \times 3 \times 2 \times 1} \\ \\ \to \: 3003$

Hope it helps you .