Question

Find the middle term in the expansion of (x/a+a/x)^12

Answer 1

$Given \: Binomial \: \big( \frac{x}{a}+ \frac{a}{x}\big)^{12}$

$Here, n = 12$

$Number \: of \: terms \: in \:the \: expansion = n+1 \\= 12 + 1 \\= 13 \: (Odd)$

$Middle \: term \: in \: the \: expansion \\= \big(\frac{n+1}{2}\big)^{th} \:term \\= \big( \frac{13+1}{2}\big)^{th} \:term \\=\big( \frac{14}{2}\big)^{th} \:term \\= 7^{th} \:term$

$\boxed { \pink { General \:term (t_{r+1})= ^{n}C_{r} \times x^{n-r} \times a^{r}}}$

$\implies Middle \:term (t_{6+1}) \\= ^{12}C_{6} \times \big(\frac{x}{a}\big)^{12-6} \times \big(\frac{a}{x}\big)^{6}$

$= \frac{12!}{ ( 12-6)! 6! } \times \big(\frac{x}{a}\big)^{6} \times \big(\frac{a}{x}\big)^{6}$

$= \frac{12!}{ 6! 6! }\\= \frac{12\times 11\times10\times 9\times8\times 7\times 6! }{(6\times 5\times4\times 3\times2\times 1)\times 6!}\\= \frac{308}{3}$

Therefore.,

$\red { Middle \:term \:in \:the \: Expansion} \green { = \frac{308}{3}}$

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