Question

Find the middle terms in the expansions of
i) ${(3-\frac{x^3}{6})}^7$
ii) ${(\frac{x}{3}+9y)}^{10}$

Answer 1

$i)$

${(3-\frac{x^3}{6})}^7$

$n=7\Rightarrow$ middle terms are $T_4,T_5$

$T_{r+1}{=}^n{C}_r{a}^{n-r}{b}^r$

$T_4=T_{3+1}{=}^7{C}_3{3}^{7-3}{(-\frac{x^3}{6})}^3$

$=-\frac{7\times 6\times 5}{3\times 2}\cdot 3\cdot \frac{x^9}{8}$

$\therefore T_4=-\frac{105}{8}{x}^9$

$T_5=T_{4+1}{=}^7{C}_4{3}^{7-4}{(-\frac{x^3}{6})}^4$

$=-\frac{7\times 6\times 5}{3\times 2}\cdot 1\cdot \frac{x^{12}}{16}$

$\therefore T_5=\frac{35}{48}{x}^{12}$

$ii)$

${(\frac{x}{3}+9y)}^{10}$

$n=10\Rightarrow$ middle term $=\frac{n}{2}+1=6$

$T_{r+1}{=}^n{C}_r{a}^{n-r}{b}^r$

$T_{5+1}{=}^{10}{C}_5{\left(\frac{x}{3}\right)}^{10-5}(9y{)}^5$

$=\frac{10\times 9\times 8\times 7\times 6\times 3^5}{5\times 4\times 3\times 2\times 1}\cdot x^5\cdot y^5$

$T_6=61236x^5{y}^5$