Find the middle terms in the expansion of {x}{3}+9y)^{10}

It is known that in the expansion of (a+b)ⁿ in n is even, then the middle term

\(( \dfrac{n}{2}+1)^{th}\)

Therefore, the middle term in the expansion of

\( ( \dfrac{10}{2}+1)^{th}\) = 6^th term

\( T_4=T_{5+1}\)

= \( ^{10}C_5(\dfrac{x}{3})^{10-5}\)

= \( (9y)^5=\dfrac{10!}{5!5!}\times \dfrac{x^5}{3^5}\times 9^5\times y^5 \)

\( \dfrac{10\times 9\times 8\times 7\times 6\times 5!}{5\times 4\times 3\times 2\times 5!}\)

\( \times \dfrac{1}{3^5}\times 3^{10}\times x^5\times y^5\)

= \( 252\times 3^5\times x^5\times y^5\)

\( =61236x^5y^5\)

Thus, the middle term in the expansion of

\( =61236x^5y^5\)